a) To create a comparative boxplot for ZOD (Zone Out Duration) by pie type, we would separate the ZOD data into two groups: apple pie and cherry pie.
The boxplots will provide a visual comparison of the distributions for the two groups. By examining the boxplots, we can determine if there appears to be a difference between the ZODs for the two groups. b) Using the set.seed(12) command to ensure reproducibility, we can create 1000 permutations of the difference in mean ZOD for cherry pie minus the mean ZOD for apple pie. The observed difference in means for the sample data would be the actual difference between the mean ZODs of the cherry pie and apple pie groups.
c) The statistical hypotheses for testing that the mean ZOD for cherry pie is greater than the mean ZOD for apple pie can be expressed as: Null hypothesis (H0): The mean ZOD for cherry pie is less than or equal to the mean ZOD for apple pie. (μcherry ≤ μapple). Alternative hypothesis (HA): The mean ZOD for cherry pie is greater than the mean ZOD for apple pie. (μcherry > μapple). d) To visualize the null distribution, we would make a histogram using the 1000 permutations. The number of bins would be set to 13. The null distribution represents the distribution of the differences in means under the assumption that there is no difference between cherry pie and apple pie ZODs. We would add a vertical line to indicate the observed sample difference in means.
e) By conducting the permutation test and setting up the code correctly, we can calculate the p-value. If the code is correct, the p-value obtained should be 0.002. This p-value represents the probability of observing a difference in means as extreme as the one observed in the sample data, assuming that there is no actual difference between cherry pie and apple pie ZODs. f) Based on the hypothesis test and the obtained p-value of 0.002, we can conclude that there is strong evidence to reject the null hypothesis. The results suggest that the mean ZOD for cherry pie is significantly greater than the mean ZOD for apple pie. In the context of the problem, it indicates that consuming cherry pie at lunch, two hours before class, leads to a higher Zone Out Duration during the lecture compared to consuming apple pie.
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abby is comparing monthly phone charges from two companies. phenix charges $30 plus $.5 per minute. Nuphone charges $40 plus $.10 per minute. in how many minutes will the total be the same
Answer:
In 25 minutes, the monthly phone charges of both companies will be the same.
Step-by-step explanation:
If we allow m to represent the number of minutes, we can create two equations for C, the total cost of phone charges from both companies:
Phoenix equation: C = 0.5m + 30
Nuphone equation: C - 0.10m + 40
Now, we can set the two equations equal to each other. Solving for m will show us how many minutes must Abby use for the total cost at both companies to be the same:
0.5m + 30 = 0.10m + 40
Step 1: Subtract 30 from both sides:
(0.5m + 30 = 0.10m + 40) - 30
0.5m = 0.10m + 10
Step 2: Subtract 0.10m from both sides:
(0.5m = 0.10m + 10) - 0.10m
0.4m = 10
Step 3: Divide both sides by 0.4 to solve for m (the number of minutes it takes for the total cost of both companies to be the same)
(0.4m = 10) / 0.4
m = 25
Thus, Abby would need to use 25 minutes for the total cost at both companies to be the same.
Optional Step 4: Check the validity of the answer by plugging in 25 for m in both equations and seeing if we get the same answer:
Checking m = 25 with Phoenix equation:
C = 0.5(25) + 30
C = 12.5 + 30
C = 42.5
Checking m = 25 with Nuphone equation:
C = 0.10(25) + 40
C = 2.5 + 40
C = 42.5
Thus, m = 25 is the correct answer.
The following questions relate to the below information.
XY2 → X + Y2
The equation above represents the decomposition of a compound XY2. The diagram below shows two reaction profiles (path one and path two) for the decomposition of XY2.
The diagram shows two reaction profiles for the decomposition of XY2, with path one representing a single-step decomposition and path two representing a two-step decomposition process.
In path one, XY2 directly decomposes into X and Y2 in a single step. This means that the decomposition reaction occurs in a single transition state without any intermediate species.
In path two, XY2 first undergoes an intermediate step where it forms an intermediate species, XY. Then, in the second step, the intermediate species XY further decomposes into X and Y. This two-step process involves two transition states.
The choice between path one and path two depends on the reaction conditions and the energy requirements for each pathway. The reaction profile diagrams provide information about the energy changes during the decomposition process.
By analyzing the reaction profiles, one can determine the activation energy required for each step and the overall energy change during the decomposition of XY2. This information is crucial for understanding the reaction kinetics and the thermodynamics of the decomposition process.
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Calculate Laplace transform of the below: 0,5 < 0 The Impulse Response: u(t) = 300,t = 0 0,t> 0 - The unit step function: u(t) = 1,t > 0 - The unit ramp function (slope=1): r(t) = t, t > 0 The exponential function: f(t) = e-atu(t),t 20 # Cosine function: f(t) = cos(wt)u(t),t>=0.
1) The Laplace transform of a function f(t) is ∫[0 to ∞] e^(-st) * f(t) dt
2) Impulse Response = 1/s
3) Unit Step Function = 1/s
4) Unit Ramp Function = 1/s^2
5) The exponential function= 1/(s + a)
6) Cosine function = -s / (s^2 + w^2),
1) The Laplace transform of a function f(t) is defined as:
F(s) = L{f(t)} = ∫[0 to ∞] e^(-st) * f(t) dt,
where s is the complex frequency parameter.
2) Impulse Response:
The impulse response u(t) can be represented as a unit step function. Therefore, the Laplace transform of the impulse response is:
L{u(t)} = ∫[0 to ∞] e^(-st) * u(t) dt
= ∫[0 to ∞] e^(-st) * 1 dt
= ∫[0 to ∞] e^(-st) dt
= [-1/s * e^(-st)] [0 to ∞]
= -1/s * (e^(-s * ∞) - e^(-s * 0))
= -1/s * (0 - 1)
= 1/s,
where s > 0.
3) Unit Step Function:
The unit step function u(t) can be directly transformed using the definition of the Laplace transform:
L{u(t)} = ∫[0 to ∞] e^(-st) * u(t) dt
= ∫[0 to ∞] e^(-st) * 1 dt
= ∫[0 to ∞] e^(-st) dt
= [-1/s * e^(-st)] [0 to ∞]
= -1/s * (e^(-s * ∞) - e^(-s * 0))
= -1/s * (0 - 1)
= 1/s,
where s > 0.
4) Unit Ramp Function:
The unit ramp function r(t) = t can be transformed as follows:
L{r(t)} = ∫[0 to ∞] e^(-st) * r(t) dt
= ∫[0 to ∞] e^(-st) * t dt
= ∫[0 to ∞] t * e^(-st) dt.
To calculate this integral, we can use integration by parts. Let's assume u = t and dv = e^(-st) dt. Then, du = dt and v = (-1/s) * e^(-st). Applying integration by parts, we have:
∫[0 to ∞] t * e^(-st) dt = [-t * (1/s) * e^(-st)] [0 to ∞] - ∫[0 to ∞] (-1/s) * e^(-st) dt
= [(-t/s) * e^(-st)] [0 to ∞] + (1/s) * ∫[0 to ∞] e^(-st) dt
= [(-t/s) * e^(-st)] [0 to ∞] + (1/s) * (1/s),
where s > 0.
Since the term (-t/s) * e^(-st) approaches zero as t approaches infinity, the first part of the integral becomes zero. Therefore, we are left with:
L{r(t)} = (1/s) * (1/s)
= 1/s^2,
where s > 0.
5) Exponential Function:
The exponential function f(t) = e^(-at) * u(t) can be transformed as follows:
L{e^(-at) * u(t)} = ∫[0 to ∞] e^(-st) * e^(-at) * u(t) dt
= ∫[0 to ∞] e^(-st - at) dt
= ∫[0 to ∞] e^(-(s + a)t) dt
= [-1/(s + a) * e^(-(s + a)t)] [0 to ∞]
= -1/(s + a) * (e^(-(s + a) * ∞) - e^(-(s + a) * 0))
= -1/(s + a) * (0 - 1)
= 1/(s + a),
where s + a > 0.
6) Cosine Function:
The cosine function f(t) = cos(wt) * u(t) can be transformed as follows:
L{cos(wt) * u(t)} = ∫[0 to ∞] e^(-st) * cos(wt) * u(t) dt
= ∫[0 to ∞] e^(-st) * cos(wt) dt.
To evaluate this integral, we can use the Laplace transform of the cosine function, which is given by:
L{cos(wt)} = s / (s^2 + w^2), where s > 0.
Therefore, we have:
L{cos(wt) * u(t)} = ∫[0 to ∞] e^(-st) * (s / (s^2 + w^2)) dt
= (s / (s^2 + w^2)) * ∫[0 to ∞] e^(-st) dt
= (s / (s^2 + w^2)) * (-1/s * e^(-st)) [0 to ∞]
= (s / (s^2 + w^2)) * (0 - 1)
= -s / (s^2 + w^2),
where s > 0.
These are the Laplace transforms of the given functions.
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what is the recursive rule for the sequence? −22.7, −18.4, −14.1, −9.8, −5.5, ...
The recursive rule for the sequence −22.7, −18.4, −14.1, −9.8, −5.5, ... is:
a(n) = a(n - 1) + 4.3
where a(n) is the nth term of the sequence.
The recursive rule for a sequence tells us how to find the next term in the sequence, given the previous terms. In this case, the recursive rule tells us that to find the next term in the sequence, we add 4.3 to the previous term.
For example, the second term in the sequence is −18.4, which is found by adding 4.3 to the first term, −22.7. The third term in the sequence is −14.1, which is found by adding 4.3 to the second term, −18.4. And so on.
The recursive rule can also be used to prove that the sequence is arithmetic.
An arithmetic sequence is a sequence of numbers in which the difference between any two consecutive terms is constant. In this case, the difference between any two consecutive terms is 4.3, so the sequence is arithmetic.
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Show that for every metric space (X, d), every x e X and every e > 0 we have: (a) CI(B.(x)) {ye X: d(x,y)
We have shown that for any metric space (X, d), any point x ∈ X, and any positive ε > 0, the open ball B(x, ε) is entirely contained within the closed ball CI(B(x)).
To prove the statement, let's consider a metric space (X, d), an arbitrary point x ∈ X, and a positive real number ε > 0. We want to show that the open ball B(x) centered at x with radius ε, denoted as B(x, ε), is contained within the closed ball CI(B(x)).
First, let y be any point in the open ball B(x, ε). This means that d(x, y) < ε, indicating that the distance between x and y is less than ε. By definition, the closed ball CI(B(x)) includes all points y in X such that d(x, y) ≤ ε. Since d(x, y) < ε implies d(x, y) ≤ ε, we can conclude that every point in the open ball B(x, ε) is also in the closed ball CI(B(x)).
Therefore, we have shown that for any metric space (X, d), any point x ∈ X, and any positive ε > 0, the open ball B(x, ε) is entirely contained within the closed ball CI(B(x)).
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A simple random sample of size n = 64 is obtained from a population with p = 76 and o=8. Describe the sampling distribution of x. (a) What is P (x>78) ? (b) What is P (x>73.6)?
a. Describing the sampling we get: P (x > 78) = 0.0228 or 2.28%.
b. The probability P (x > 73.6) = 0.9953 or 99.53%.
The sampling distribution of x is normally distributed with a mean of µ = 76 and a standard deviation of σ = 8/√64 = 1. (a) The z-score for a sample mean of x > 78 is (78 - 76) / (8 / √64) = 2. The probability of a z-score greater than 2 is approximately 0.0228 or 2.28%. Hence P (x > 78) = 0.0228 or 2.28%.
(b) The z-score for a sample mean of x > 73.6 is (73.6 - 76) / (8 / √64) = -2.6. The probability of a z-score greater than -2.6 is approximately 0.9953 or 99.53%. Hence P (x > 73.6) = 0.9953 or 99.53%.
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A simple random sample of size n = 64 is obtained from a population with p = 76 and o=8.
We have to find the following inferences from the sample statistics
Mean of the sampling distribution of x is μx=μ=76 (the population mean).
Standard deviation of the sampling distribution of x is σx=σ/√n=8/√64=1
Shape of the distribution is approximately normal by the central limit theorem.
Now we know that standard normal variate is calculated as:
z= x - μx/σx = x - μ / σx
where x is the random variable.P (x>78) is to be calculated.
Using the above formula, we get:
[tex]P (x>78) = P(z>78 - 76 / 1)= P(z>2)[/tex]
At z=2, the area is 0.0228.
Hence,P (x>78) = P(z>2)= 0.0228 (approximately)
Using the above formula, we get:
[tex]P (x>73.6) = P(z>73.6 - 76 / 1)= P(z>-2.4)[/tex]
At z=-2.4, the area is 0.0082.
Hence,[tex]P (x>73.6) = P(z>-2.4)= 0.0082[/tex] (approximately)
Therefore, the answers are:(a) P (x>78) = 0.0228 (approximately)(b) P (x>73.6) = 0.0082 (approximately).
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Given f(x)=11^x, what is f^-1(x)?
Answer:
The first one
[tex] log_{11} \: (x)[/tex]
Step-by-step explanation:
f(x) = 11^x
Here are the steps to find the inverse of a function:
1. Let f(x)=y
2. Make x the subject of formula.
3. Replace y by x.
[tex]11 {}^{x} = y \\ \: log(11 {}^{x} ) = log(y) \\ x log(11) = log(y) \\ x = \frac{ log(y) }{ log(11) } = log_{11}(y) \\ f {}^{ - 1} (x) = log_{11}(x) [/tex]
A sample of 20 from a population produced a mean of 66.0 and a standard deviation of 10.0. A sample of 25 from another population produced a mean of 58.6 and a standard deviation of 13.0. Assume that the two populations are normally distributed and the standard deviations of the two populations are equal. The null hypothesis is that the two population means are equal, while the alternative hypothesis is that the two population means are different. The significance level is 5%.
What is the value of the test statistic, t, rounded to three decimal places?
Type your answer here
The value of the test statistic (t) is approximately 2.157.
Formula for test statistic?
To calculate the test statistic (t), we can use the formula:
[tex]t = (x_1 - x_2) / \sqrt{(s_1^2 / n_1) + (s_2^2 / n_2)}[/tex]
Where:
[tex]x_1[/tex] and [tex]x_2[/tex] are the sample means,
[tex]s_1[/tex] and [tex]s_2[/tex] are the sample standard deviations,
[tex]n_1[/tex] and [tex]n_2[/tex] are the sample sizes.
Given:
[tex]x_1[/tex] = 66.0, [tex]x_2[/tex] = 58.6,
[tex]s_1[/tex] = 10.0, [tex]s_2[/tex] = 13.0,
[tex]n_1[/tex] = 20, [tex]n_2[/tex] = 25.
Substituting the values into the formula, we have:
[tex]t = (66.0 - 58.6) / \sqrt{(10.0^2 / 20) + (13.0^2 / 25)}[/tex]
Calculating the expression in the square root first:
[tex]t = (66.0 - 58.6) / \sqrt{(5.0) + (6.76)}[/tex]
[tex]t = 7.4 / \sqrt{(11.76)}[/tex]
Finally, calculating the square root and dividing:
t ≈ 7.4 / 3.429
t ≈ 2.157
Rounding to three decimal places, the value of the test statistic (t) is approximately 2.157.
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The lifetime of a certain bulb is exponential with a mean of 3 years. If we take a random sample of 10 such bulbs, what is the expected number of bulbs which will last at least 1 year? What is the probability that exactly 4 of the 10 bulbs will last at least 1 year?
The probability that exactly 4 of the 10 bulbs will last at least 1 year ≈ 0.2405 or 24.05%.
The lifetime of a certain bulb is exponentially distributed with a mean of 3 years. This means that the rate parameter (λ) of the exponential distribution is equal to 1/3.
To find the expected number of bulbs that will last at least 1 year, we can use the exponential distribution's cumulative distribution function (CDF).
The CDF of an exponential distribution is given by:
CDF(x) = 1 - exp(-λx)
To find the probability that a bulb will last at least 1 year, we calculate the CDF at x = 1:
CDF(1) = 1 - exp(-1/3 * 1) = 1 - exp(-1/3) ≈ 0.2835
Therefore, the expected number of bulbs that will last at least 1 year in a sample of 10 bulbs is:
Expected number = 10 * CDF(1) = 10 * 0.2835 = 2.835 bulbs
To find the probability that exactly 4 of the 10 bulbs will last at least 1 year, we can use the binomial distribution.
The probability mass function (PMF) of the binomial distribution is given by:
PMF(k) = (n choose k) * p^k * (1-p)^(n-k)
where n is the number of trials, k is the number of successful trials, and p is the probability of success in a single trial.
In this case, n = 10, k = 4, and p = CDF(1) ≈ 0.2835.
Plugging these values into the PMF formula, we get:
PMF(4) = (10 choose 4) * (0.2835)^4 * (1 - 0.2835)^(10-4)
Using a binomial coefficient calculator, we find:
(10 choose 4) = 210
Calculating the probability:
PMF(4) = 210 * (0.2835)^4 * (1 - 0.2835)^6 ≈ 0.2405
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The student council at a large high school is wondering if Juniors or Seniors are more likely to attend Prom. They take a random sample of 40 Juniors and find that 18 are planning on attending Prom. They select a random sample of 38 Seniors and 19 are planning on attending. Do the data provide convincing evidence that a higher proportion of Seniors are going to prom than Juniors? Use a 5% significance level. What is the p-value? Round to two decimal places. O 0.33 0.21 O 0.56
The data provide convincing evidence that a higher proportion of Seniors are attending prom compared to Juniors. The p-value is 0.33.
To determine if a higher proportion of Seniors are attending prom compared to Juniors, we can conduct a hypothesis test using the given data. Let's set up the hypotheses:
Null hypothesis (H0): The proportion of Juniors attending prom is equal to or higher than the proportion of Seniors attending prom.
Alternative hypothesis (Ha): The proportion of Seniors attending prom is higher than the proportion of Juniors attending prom.
To test this, we can use a two-sample proportion z-test. First, let's calculate the proportions of Juniors and Seniors attending prom:
Proportion of Juniors attending prom: 18/40 = 0.45
Proportion of Seniors attending prom: 19/38 = 0.50
Next, we calculate the standard error of the difference in proportions:
SE = [tex]\sqrt{[(p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2)]}[/tex]
SE = [tex]\sqrt{[(0.45 * 0.55 / 40) + (0.50 * 0.50 / 38)]}[/tex]
SE ≈ 0.090
We can now calculate the test statistic (z-score):
z = (p1 - p2) / SE
z = (0.45 - 0.50) / 0.090
z ≈ -0.56
Looking up the z-score in the z-table, we find that the p-value associated with -0.56 is approximately 0.33. Since the p-value (0.33) is greater than the significance level of 0.05, we fail to reject the null hypothesis. Therefore, we do not have convincing evidence to conclude that a higher proportion of Seniors are attending prom compared to Juniors.
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Consider a single server queue with a Poisson arrival process at rate \, and exponentially distributed service times with rate µ. All interarrival times and service times are independent of each other. This is similar to the standard M|M|1 queue, but in this queue, as the queue size increases, arrivals are more and more likely to decide not to join it. If an arrival finds n people already in the queue ahead of them (including anyone being served), then they join with probability 1/(n+1). Let N(t) be the number in the queue at time t. (c) Find the equilibrium distribution for this queue, when it exists. (d) What are conditions on A and under which the equilibrium distribution exists?
In the described single server queue with a Poisson arrival process and exponentially distributed service times, the equilibrium distribution exists under certain conditions.
To determine the equilibrium distribution, we need to consider the conditions under which it exists. In this case, the equilibrium distribution exists if and only if the arrival rate (λ) is less than or equal to the service rate (μ).
Mathematically, λ ≤ μ. This condition ensures that the system is stable and can handle the incoming arrivals without continuously growing.
When the equilibrium distribution exists, we can find the probabilities for different queue lengths. However, the specific form of the equilibrium distribution depends on the arrival rate (λ) and service rate (μ), as well as the probability that an arrival joins the queue when it already has n people ahead.
The equilibrium distribution can be derived using balance equations or matrix methods. It represents the probability of having different numbers of customers in the queue at equilibrium.
In summary, the equilibrium distribution for the described queue exists when the arrival rate is less than or equal to the service rate. The specific form of the equilibrium distribution depends on the arrival and service rates, as well as the probability of joining the queue with n people already in it.
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(f) Another river is a smaller but very important source of water flowing out of the park from a different drainage. Ten recent years of annual water flow data are shown below (units 10^8 cubic meters).
3.83 3.81 4.01 4.84 5.81 5.50 4.31 5.81 4.31 4.57
Although smaller, is the new river more reliable? Use the coefficient of variation to make an estimate. (Round your answers to two decimal place.)
original river's coefficient of variation ____
smaller river's coefficient of variation ____
What do you conclude?
A. The smaller river is more consistent.
B. Neither river is more consistent.
C. The original river is more consistent.
(g) Based on the data, would it be safe to allocate at least 26 units of the orginal river water each year for agricultural and domestic use? Why or why not?
A. No, the median is less than 26 which means more than half the river flows are below 26.
B. No, Q3 is less than 26 which means more than three quarters of the river flows are below 26.
C. No, since 26 is an upper outlier it will be very rare to have a flow at or above 26.
D. Yes, since 26 is an lower outlier it will be very rare to have a flow below 26.
E. Yes, Q1 is greater than 26 which means over three quarters of the river flows are at or above 26.
The correct answer is option A: No, the median is less than 26 which means more than half the river flows are below 26 based on coefficient of variation.
The smaller river's coefficient of variation can be calculated as shown below;
Small river's mean=4.5
Standard deviation
=√( (3.83-4.5)²+(3.81-4.5)²+(4.01-4.5)²+(4.84-4.5)²+(5.81-4.5)²+(5.50-4.5)²+(4.31-4.5)²+(5.81-4.5)²+(4.31-4.5)²+(4.57-4.5)² )/(10-1)
≈0.67
Coefficient of variation= (0.67/4.5)*100
= 14.89%
Original river's coefficient of variation can be calculated as shown below:
Original river's mean=16.5
Standard deviation
=√( (18.3-16.5)²+(17.5-16.5)²+(14.9-16.5)²+(21.3-16.5)²+(15.3-16.5)²+(13.1-16.5)²+(19.6-16.5)²+(14.7-16.5)²+(15.6-16.5)²+(14.6-16.5)² )/(10-1)
≈2.21
Coefficient of variation= (2.21/16.5)*100
= 13.39%
Hence the coefficient of variation for the smaller river is greater than that of the original river.
Thus, we can conclude that the original river is more consistent.
Safe allocation of water 26 is greater than the Q1 of the original river, which implies that the lower 25% of the river flows are less than 26 units.
Therefore, it is not safe to allocate at least 26 units of the original river water each year for agricultural and domestic use.
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A radio station surveyed 195 students to determine the sports they liked. They found 70 liked football, 95 liked shuffleboard, and 60 liked neither type. Let U = {all students surveyed}, F = {students who liked football}, S = {students who liked shuffleboard}. How many of the students liked at least one of the two sports?
A radio station surveyed 195 students out of which 75 of the students liked at least one of the two sports.
In this question, we are given three sets of data related to students of a radio station. We have to find out the number of students who liked at least one of the two sports.
Let U = All students surveyed F = Students who liked football S = Students who liked shuffleboard
The formula we are going to use in this question is given below
n(F ∪ S) = n(F) + n(S) - n(F ∩ S)
Where ∪ represents union, ∩ represents intersection, n represents the number of elements in the set and the total number of students surveyed is U = 195.
The information given in the question is represented in the Venn diagram below: Venn diagram of the information given in the question
We have to find out the number of students who liked at least one of the two sports.
To find this, we need to add the number of students who liked football to the number of students who liked shuffleboard and then subtract the number of students who liked both sports (intersection of F and S).
n(F ∪ S) = n(F) + n(S) - n(F ∩ S)n(F ∪ S) = 70 + 95 - n(F ∩ S)
Now we have to find the number of students who liked both sports.
According to the information given in the question:
n(U) = 195 n(F) = 70 n(S) = 95
n(U − F − S) = 60
n(F ∩ S) = ?
We can calculate n(F ∩ S) as follows:
n(U − F − S) = 60
n(F ∩ S) = n(U) − n(F) − n(S) + n(F ∩ S)
n(F ∩ S) = 195 - 70 - 95 + 60 = 90
Now we can substitute the values of n(F) = 70, n(S) = 95, and n(F ∩ S) = 90 in the formula:
n(F ∪ S) = n(F) + n(S) - n(F ∩ S)
n(F ∪ S) = 70 + 95 - 90n(F ∪ S) = 75
Therefore, the number of students who liked at least one of the two sports is 75.
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Find the average value of the function over the given interval. (Round your answer to four decimal places.) f(x) = 4 – x², [-2, 2]
The average value of the function f(x) = 4 - x² over the interval [-2, 2] is 4.
To find the average value of the function f(x) = 4 - x² over the interval [-2, 2], we need to evaluate the definite integral of the function over that interval and divide it by the width of the interval.
The average value of f(x) over the interval [a, b] is given by the formula:
Average value = (1 / (b - a)) * ∫[a to b] f(x) dx
In this case, a = -2 and b = 2. Let's calculate the average value using the formula:
Average value = (1 / (2 - (-2))) * ∫[-2 to 2] (4 - x²) dx
First, we integrate the function:
∫(4 - x²) dx = [4x - (x³ / 3)] evaluated from -2 to 2
Plugging in the limits:
[4(2) - ((2³) / 3)] - [4(-2) - ((-2³) / 3)]
Simplifying further:
[8 - (8 / 3)] - [-8 - (8 / 3)]
Combining like terms:
[24 / 3 - 8 / 3] - [-24 / 3 - 8 / 3]
(16 / 3) - (-32 / 3) = 48 / 3 = 16
Now, we divide the result by the width of the interval:
Average value = 16 / (2 - (-2)) = 16 / 4 = 4
Therefore, the average value of the function f(x) = 4 - x² over the interval [-2, 2] is 4.
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The time doctors spend with patients is normally distributed with a mean of 21.6 minutes and a standard deviation of 1.8 minutes. The slowest 18% of doctors will spend more than how many minutes with patients? (2 decimal places)
The slowest 18% of doctors will spend more than 19.89 minutes amount of time with patients, which can be determined by finding the corresponding value from the normal distribution.
Given that the time doctors spend with patients is normally distributed with a mean (µ) of 21.6 minutes and a standard deviation (σ) of 1.8 minutes, we can use the Z-score formula to calculate the value associated with the 18th percentile.
Step 1: Convert the percentile to a Z-score
Z = InvNorm(0.18) = -0.9154 (using a standard normal distribution table or calculator)
Step 2: Calculate the value associated with the Z-score
X = µ + Z * σ
X = 21.6 + (-0.9154) * 1.8
X ≈ 19.89
Therefore, the slowest 18% of doctors will spend more than approximately 19.89 minutes with patients.
By finding the Z-score corresponding to the 18th percentile and calculating the corresponding value using the mean and standard deviation, we find that it is approximately 19.89 minutes.
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Use appropriate algebra and Theorem 7.2.1 to find the given inverse Laplace transform. (Write your answer as a function of t.) L-1 (4s/4s^2+1)
The inverse Laplace transform of [tex]L^{-1}[/tex](4s/(4s² + 1)) is [tex]e^{(-i/2t)[/tex] + [tex]e^{(i/2t))/2[/tex].
To find the inverse Laplace transform of [tex]L^{-1}[/tex](4s/(4s² + 1)), we can use partial fraction decomposition.
Step 1: Factorize the denominator of the Laplace transform.
4s² + 1 = (2s + i)(2s - i)
Step 2: Write the partial fraction decomposition.
4s/(4s² + 1) = A/(2s + i) + B/(2s - i)
Step 3: Clear the fractions.
4s = A(2s - i) + B(2s + i)
Step 4: Solve for A and B.
Comparing coefficients:
4 = 2A + 2B (coefficient of s terms)
0 = -Ai + Bi (constant terms)
From the second equation, we can see that A = B. Substituting this into the first equation:
4 = 4A
A = 1
So, B = 1 as well.
Step 5: Rewrite the partial fraction decomposition.
4s/(4s² + 1) = 1/(2s + i) + 1/(2s - i)
Step 6: Take the inverse Laplace transform.
[tex]L^{-1}[/tex](4s/(4s² + 1)) = [tex]L^{-1}[/tex](1/(2s + i)) + [tex]L^{-1}[/tex](1/(2s - i))
Using Theorem 7.2.1, the inverse Laplace transforms of the individual terms can be found:
[tex]L^{-1}[/tex](1/(2s + i)) = [tex]e^{(-i/2t)/2[/tex]
[tex]L^{-1}[/tex](1/(2s - i)) = [tex]e^{(i/2t)/2[/tex]
Therefore, the inverse Laplace transform of [tex]L^{-1}[/tex](4s/(4s² + 1)) is:
[tex]L^{-1}[/tex](4s/(4s² + 1)) = [tex]e^{(-i/2t)/2[/tex] + [tex]e^{(i/2t)/2[/tex].
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For the upcoming 2024 presidential election, Donald Trump represents the republican party and Joe Biden represents the democratic party. A third candidate Ashley Tisdale represents the independent party. The probabilities that a registered voter voters for Trump, Biden and Tisdale are Pp_1, p_2 and p_3, respectively. Out of a random sample of 10,000 voters, it is found that 4800 voted for Trump, 4400 voted for Biden and 800 voted for Tisdale.
(a) Find an approximate 98% lower confidence interval for p_1 – p_2.
(b) Based on (a), is there any convincing evidence that Trump will win the election?
HINT: You have to estimate the variance of p_1 – p_2 using the given data and then apply the bivariate version of the Central Limit The- orem. You must understand the difference between this experiment and rolling two dice independently.
The approximate 98% lower confidence interval for p₁ - p₂ is (0.003328, 0.076672).
Based on the value of p₁ - p₂, there is convincing evidence that Trump will win the election.
What is the confidence interval?(a) To find an approximate 98% lower confidence interval for p₁ - p₂, we can use the following formula:
CI = (p₁ - p₂) ± z * √((p₁ * (1 - p₁) / n₁) + (p₂ * (1 - p₂) / n₂))
where:
p₁ and p₂ are the sample proportions (p₁ = 4800/10000, p₂ = 4400/10000),
n₁ and n₂ are the respective sample sizes (n₁ = 10000, n₂ = 10000),
z is the z-score (98% confidence level corresponds to a z-score of 2.33).
Substituting the values into the formula:
CI = (0.48 - 0.44) ± 2.33 * √((0.48 * 0.52 / 10000) + (0.44 * 0.56 / 10000))
CI = 0.04 ± 2.33 * √(0.0001248 + 0.0001232)
CI = 0.04 ± 2.33 * √(0.000248)
CI = 0.04 ± 2.33 * 0.0157496
CI ≈ 0.04 ± 0.036672
CI ≈ (0.003328, 0.076672)
(b) The lower bound of the interval is greater than zero (0.003328 > 0), therefore, based on the confidence interval, there is convincing evidence that the proportion of voters supporting Trump (p₁) is higher than the proportion of voters supporting Biden (p₂).
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1) If total costs for a product are given by C(x) = 1760 + 8x + 0.6x2 and total revenues are given by R(x) = 100x -0.4x2, find the break-even points. =
2) If total costs for a commodity are given by C(x) = 900 +25x and total revenues are given by R(x) = 100x - x2, find the break-even points. 3) Find the maximum revenue and maximum profit for the functions described in Problem #2.
a) The break-even points for the given cost and revenue functions are approximately x = 16.526 and x = 6.474.
b) The break-even points for the given cost and revenue functions are approximately x = 12.225 and x = 62.775.
c) The maximum profit for the given cost and revenue functions is approximately $843.75.
a) To find the break-even points, we need to determine the values of x where the total costs (C(x)) equal the total revenues (R(x)). We set C(x) = R(x) and solve for x:
C(x) = R(x)
1760 + 8x + 0.6x² = 100x - 0.4x²
Combining like terms and rearranging the equation, we get:
1x² - 92x + 1760 = 0
Solving this quadratic equation, we find two solutions for x:
x ≈ 16.526
x ≈ 6.474
b) Similarly, we set C(x) = R(x) and solve for x:
900 + 25x = 100x - x²
Rearranging the equation, we get:
x² - 75x + 900 = 0
Solving this quadratic equation, we find two solutions for x:
x ≈ 12.225
x ≈ 62.775
c) To find the maximum revenue, we need to determine the vertex of the revenue function R(x) = 100x - x². The x-coordinate of the vertex is given by x = -b / (2a), where a and b are the coefficients of the quadratic equation.
In this case, a = -1 and b = 100. Plugging in the values, we get:
x = -100 / (2 * -1) = 50
Substituting this value back into the revenue function, we find:
R(50) = 100(50) - (50)² = 5000 - 2500 = 2500
Therefore, the maximum revenue for the given cost and revenue functions is $2500.
To find the maximum profit, we need to subtract the total costs from the total revenues. Given that the cost function is C(x) = 900 + 25x, the profit function is P(x) = R(x) - C(x). Substituting the revenue and cost functions, we have:
P(x) = (100x - x²) - (900 + 25x)
P(x) = -x² + 75x - 900
To find the maximum profit, we need to determine the vertex of the profit function. Using the same formula as before, we find:
x = -75 / (2 * -1) = 37.5
Substituting this value back into the profit function, we find:
P(37.5) = -(37.5)² + 75(37.5) - 900 ≈ 843.75
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Tickets for a raffle cost $$8. There were 734 tickets sold. One ticket will be randomly selected as the winner, and that person wins $$1500 and also the person is given back the cost of the ticket. For someone who buys a ticket, what is the Expected Value (the mean of the distribution)?
If the Expected Value is negative, be sure to include the "-" sign with the answer. Express the answer rounded to two decimal places.
2, A ping pong ball is drawn at random from an urn consisting of balls numbered 2 through 10. A player wins 1 dollar if the number on the ball is odd and loses 1 dollar if the number is even. What is the expected value of his winnings? Express your answer in fraction form.
3,A card is drawn at random from a standard deck of playing cards (no jokers). If it is red, the player wins 1 dollar; if it is black, the player loses 2 dollars. Find the expected value of the game. Express your answer in fraction form.
4,A bag contains 2 gold marbles, 6 silver marbles, and 29 black marbles. Someone offers to play this game: You randomly select one marble from the bag. If it is gold, you win $4. If it is silver, you win $3. If it is black, you lose $1.
What is your expected value if you play this game?
5,A bag contains 3 gold marbles, 10 silver marbles, and 30 black marbles. Someone offers to play this game: You randomly select one marble from the bag. If it is gold, you win $4. If it is silver, you win $3. If it is black, you lose $1.
What is your expected value if you play this game?
1. The expected value is then (1/734) * ($1508) + (733/734) * (-$8). 2. The expected value is (3/43) * $4 + (10/43) * $3 + (30/43) * (-$1). expected value is (5/10) * $1 + (5/10) * (-$1). 3. The expected value is (3/43) * $4 + (10/43) * $3 + (30/43) * (-$1). expected value is (26/52) * $1 + (26/52) * (-$2). 4. The expected value is (2/37) * $4 + (6/37) * $3 + (29/37) * (-$1). 5. The expected value is (3/43) * $4 + (10/43) * $3 + (30/43) * (-$1).
For the raffle ticket, the expected value is calculated by multiplying the probability of winning ($1500 + $8) by the probability of not winning (-$8). The total number of tickets sold is 734, so the probability of winning is 1/734. The expected value is then (1/734) * ($1508) + (733/734) * (-$8).
The expected value of the ping pong ball game is calculated by finding the probability of winning $1 (odd number) and losing $1 (even number) for each possible outcome (numbers 2 through 10). Since there are 5 odd numbers and 5 even numbers, the expected value is (5/10) * $1 + (5/10) * (-$1).
The expected value of the card game is calculated by finding the probability of drawing a red card and winning $1, and the probability of drawing a black card and losing $2. Since there are 26 red cards and 26 black cards in a standard deck, the expected value is (26/52) * $1 + (26/52) * (-$2).
The expected value of the marble game is calculated by multiplying the probability of drawing each type of marble (gold, silver, and black) by the corresponding amount won or lost. The expected value is (2/37) * $4 + (6/37) * $3 + (29/37) * (-$1).
Similar to the previous game, the expected value of the marble game is calculated by multiplying the probability of drawing each type of marble (gold, silver, and black) by the corresponding amount won or lost. The expected value is (3/43) * $4 + (10/43) * $3 + (30/43) * (-$1).
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The joint PDF for random variables X and Y is given as if 0 < x < 1, 0 < y < 2 x = fx.r(2, 4) = { A(48 + 3) 0.W. a) Sketch the sample space. b) Find A so that fx,y(x, y) is a valid joint pdf. c) Find the marginal PDFs fx(x) and fy(y). Are X, Y independent? d) Find P[] < X < 2,1
a) |\
| \
Y | \
| \
| \
| ____ \
X
b) A = 1/102
c) Marginal PDF fx(x) = (1/102) * x
Marginal PDF fy(y) = (51/102)
No, X and Y are not independent since their marginal PDFs fx(x) and fy(y) are not separable (i.e., they cannot be expressed as the product of individual PDFs)
d) P(0 < X < 2, 1) = 1.
a) To sketch the sample space, we need to consider the ranges of X and Y as defined in the problem statement: 0 < x < 1 and 0 < y < 2x. This means that X ranges from 0 to 1 and Y ranges from 0 to 2X. The sample space can be represented by a triangular region bounded by the lines Y = 0, X = 1, and Y = 2X.
|\
| \
Y | \
| \
| \
| ____ \
X
b) To find the value of A so that fx,y(x, y) is a valid joint PDF, we need to ensure that the joint PDF integrates to 1 over the entire sample space.
The joint PDF is given by fx,y(x, y) = A(48 + 3), where 0 < x < 1 and 0 < y < 2x.
To find A, we integrate the joint PDF over the sample space:
∫∫fx,y(x, y) dy dx = 1
∫∫A(48 + 3) dy dx = 1
A∫∫(48 + 3) dy dx = 1
A(48y + 3y)∣∣∣0∣∣2xdx = 1
A(96x + 6x)∣∣∣0∣∣1 = 1
A(96 + 6) = 1
102A = 1
A = 1/102
Therefore, A = 1/102.
c) To find the marginal PDFs fx(x) and fy(y), we integrate the joint PDF over the respective variables.
Marginal PDF fx(x):
fx(x) = ∫fy(x, y) dy
Since 0 < y < 2x, the integral limits for y are 0 to 2x.
fx(x) = ∫A(48 + 3) dy from 0 to 2x
fx(x) = A(48y + 3y)∣∣∣0∣∣2x
fx(x) = A(96x + 6x)
fx(x) = 102A * x
fx(x) = (1/102) * x
Marginal PDF fy(y):
fy(y) = ∫fx(x, y) dx
Since 0 < x < 1, the integral limits for x are 0 to 1.
fy(y) = ∫A(48 + 3) dx from 0 to 1
fy(y) = A(48x + 3x)∣∣∣0∣∣1
fy(y) = A(48 + 3)
fy(y) = A(51)
fy(y) = (51/102)
No, X and Y are not independent since their marginal PDFs fx(x) and fy(y) are not separable (i.e., they cannot be expressed as the product of individual PDFs).
d) To find P(0 < X < 2, 1), we need to integrate the joint PDF over the given range.
P(0 < X < 2, 1) = ∫∫fx,y(x, y) dy dx over the region 0 < x < 2 and 0 < y < 1
P(0 < X < 2, 1) = ∫∫A(48 + 3) dy dx over the region 0 < x < 2 and 0 < y < 1
P(0 < X < 2, 1) = A(48y + 3y)∣∣∣0∣∣1 dx over the region 0 < x < 2
P(0 < X < 2, 1) = A(48 + 3) dx over the region 0 < x < 2
P(0 < X < 2, 1) = A(51x)∣∣∣0∣∣2
P(0 < X < 2, 1) = A(102)
P(0 < X < 2, 1) = (1/102)(102)
P(0 < X < 2, 1) = 1
Therefore, P(0 < X < 2, 1) = 1.
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Please help! Due tonight.
The lateral surface area of the pyramid is 126 m².
Option C is the correct answer.
We have,
The lateral area means the surface area except for the base and the top area.
Now,
The pyramid is a triangular pyramid.
There are three faces and each face is a triangle.
Now,
Area of a triangle.
= 1/2 x base x height
= 1/2 x 7 x 12
= 7 x 6
= 42 m²
Now,
Since all three triangular faces are the same.
The lateral surface area of the pyramid.
= 3 x 42
= 126 m²
Thus,
The lateral surface area of the pyramid is 126 m².
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A pair of fair dice is tossed. Events A and B are defined as follows.
A: {The sum of the numbers on the dice is 3}
B: {At least one of the dice shows a 2}
Identify the sample points in the event A ∩ B.
The sample point in the event [tex]A \cap B[/tex] is {(2, 1)}.
To identify the sample points in an event [tex]A \cap B[/tex], we need to find the outcomes where both events A and B occur simultaneously.
Event A: The sum of the numbers on the dice is 3. The possible outcomes that satisfy this event are:
{(1, 2), (2, 1)}
Event B: At least one of the dice shows a 2. The possible outcomes that satisfy this event are:
{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
To find the sample points in the intersection of events [tex]A \cap B[/tex], we need to identify the outcomes that are common to both events. In this case, the common outcome is (2, 1).
Therefore, the sample point in the event [tex]A \cap B[/tex] is {(2, 1)}.
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Pls help ASAP! Show work
Option D is correct, the solid is a rectangular prism with a base length of 8.
The plane region is revolved completely about the x axis to sweep out a solid of revolution.
From the given figure we can tell that the solid shape obtained is a rectangular prism.
The rectangular prism has a base length of 8 units.
We have to find the volume:
volume = length × width × height
=8×5×5
=200 cubic units.
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in the coordinate plane, what is the length of the line segment that connects points at (4, −1) and (9, 7)? enter your answer in the box. round to the nearest hundredth.
The length of the line segment is approximately 9.43 units.
To find the length of the line segment connecting two points in the coordinate plane, we can use the distance formula. The distance formula calculates the distance between two points (x₁, y₁) and (x₂, y₂) as follows:
Distance = √((x₂ - x₁)² + (y₂ - y₁)²)
In this case, the coordinates of the two points are (4, -1) and (9, 7). Let's substitute these values into the distance formula:
Distance = √((9 - 4)² + (7 - (-1))²)
= √(5² + 8²)
= √(25 + 64)
= √89
≈ 9.43
Rounding to the nearest hundredth, the length of the line segment is approximately 9.43.
To justify the solution, we can visually represent the line segment connecting the two points (4, -1) and (9, 7) on a coordinate plane. By plotting these points and drawing a straight line between them, we can observe that the line segment's length corresponds to the distance between the points. We can use a ruler or any measuring tool to measure this distance on the graph, and it will match the calculated value of approximately 9.43.
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Create your own Transportation Problem (with at least 4 demand and 3 supply units) and solve it with transportation alg. (use Vogel App. Method for starting solution)
To find the total transportation cost, the allocation cost for each cell is multiplied by the unit cost, and the sum is taken. The sum of these costs is $12,800.
Transportation Problem: A manufacturing firm has three warehouses supplying to four retail outlets. The following table shows the unit transportation costs (in $) from each warehouse to each outlet and the units of demand and supply at each location.
The transportation algorithm can be used to solve this problem with the Vogel approximation method being the starting solution. Below is the transportation table (in dollars):
| | Retail Outlet 1 | Retail Outlet 2 | Retail Outlet 3 | Retail Outlet 4 | Supply |
Warehouse 1 | 6 | 5 | 3 | 7 | 300 |
Warehouse 2 | 9 | 7 | 4 | 6 | 200 |
Warehouse 3 | 2 | 8 | 5 | 9 | 250 |
Demand | 200 | 150 | 100 | 200 | |
The Vogel approximation method is an iterative procedure that selects the smallest difference between the two smallest costs for each row or column and then assigns the maximum possible allocation to it.
Step 1:
Subtract the smallest cost from the second-smallest cost and record the differences for each row and column. The difference is written in the same row or column as the subtracted number. The differences are calculated as follows:
| | Retail Outlet 1 | Retail Outlet 2 | Retail Outlet 3 | Retail Outlet 4 | Supply |
Warehouse 1 | 6 | 5 | 3 | 7 | 300 |
Warehouse 2 | 9 | 7 | 4 | 6 | 200 |
Warehouse 3 | 2 | 8 | 5 | 9 | 250 |
Demand | 200 | 150 | 100 | 200 | |
The differences are as follows:
| | Retail Outlet 1 | Retail Outlet 2 | Retail Outlet 3 | Retail Outlet 4 | Supply |
Warehouse 1 | 1 | 2 | 0 | 4 | 300 |
Warehouse 2 | 3 | 1 | 0 | 2 | 200 |
Warehouse 3 | 3 | 1 | 0 | 4 | 250 |
Demand | 200 | 150 | 100 | 200 | |
Step 2:
Identify the largest difference for each row or column and then select the smallest number in that row or column for the next allocation. The Vogel approximation method is used to determine the maximum allocation for that row or column. The total cost is then multiplied by the unit cost. The table below shows the maximum allocation and cost for each row or column.
The cost of transportation is shown below:
| | Retail Outlet 1 | Retail Outlet 2 | Retail Outlet 3 | Retail Outlet 4 | Supply |
Warehouse 1 | 6 | 5 | 3 | 7 | 300 |
Warehouse 2 | 9 | 7 | 4 | 6 | 200 |
Warehouse 3 | 2 | 8 | 5 | 9 | 250 |
Demand | 200 | 150 | 100 | 200 | |
To find the total transportation cost, the allocation cost for each cell is multiplied by the unit cost, and the sum is taken. The sum of these costs is $12,800.
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The final solution to the given transportation problem, with a minimum cost of 2050 units, is shown below:
D1 | D2 | D3 | D4 | S1 | 30 | 20 | 30 | 20 | S2 | 0 | 60 | 20 | 30 | S3 | 10 | 0 | 10 | 40 | Total Cost | 1800 | 600 | 650 | 2050 |
Explanation:
A transportation problem is one of the most fundamental optimization problems that exist. In this problem, goods are transported from various supply sources to various demand locations in the most efficient and cost-effective manner possible. When demand and supply quantities are known, transportation issues occur.
Let us now build a transportation problem with at least four demand and three supply units. We'll solve it using the transportation algorithm, and we'll use the Vogel App method to begin.
The problem is as follows:
Let us suppose that there are three factories (supply locations), S1, S2, and S3, and four warehouses (demand locations), D1, D2, D3, and D4. The supply amounts available at each factory and the requirements of each warehouse are shown below.
Supply (units) | Demand (units) | S1 | S2 | S3 | D1 | 60 | 30 | 40 | 50 | D2 | 30 | 70 | 20 | 30 | D3 | 40 | 20 | 10 | 40 | D4 | 20 | 60 | 30 | 10 |
To begin, let us generate the initial table below, which includes the amount of units available from each source to each destination.
Supply (units) | Demand (units) | S1 | S2 | S3 | Availability | D1 | 60 | 30 | 40 | 130 | D2 | 30 | 70 | 20 | 120 | D3 | 40 | 20 | 10 | 70 | D4 | 20 | 60 | 30 | 110 |
Requirement | 50 | 30 | 40 | 120 |
We'll begin by calculating the difference between the two smallest costs for each supply and demand row. Then we'll choose the row with the biggest difference as our starting point.
In this case, the differences for the supply rows are:
Supply (units) | Demand (units) | S1 | S2 | S3 | Availability | D1 | 60 | 30 | 40 | 130 | 20 | D2 | 30 | 70 | 20 | 120 | 30 | D3 | 40 | 20 | 10 | 70 | 10 | D4 | 20 | 60 | 30 | 110 | 20 |
Requirement | 50 | 30 | 40 | 120 |
Difference | 10 | 20 | 30 | |
We'll choose the third row (supply from S3) as our starting point since it has the largest difference of 30. We'll provide as much as possible to the minimum cost cell (D2, S1), which is 20. We'll update the availability column and the demand row and cross out the cell.
D1 | D2 | D3 | D4 | S1 | 40 | 0 | 40 | 20 | S2 | 30 | 70 | 20 | 30 | S3 | 0 | 0 | 0 | 50 |
Availability | 20 | 50 | 10 | 90 |
Requirement | 50 | 10 | 40 | 120 |
We'll now update the differences based on the available cells (we only have two remaining).
Supply (units) | Demand (units) | S1 | S2 | S3 | Availability | D1 | 40 | 0 | 40 | 110 | 20 | D2 | 0 | 50 | 0 | 100 | 10 | D3 | 40 | 20 | 10 | 70 | 10 | D4 | 20 | 10 | 30 | 100 | 20 |
Requirement | 50 | 20 | 40 | 120 |
Difference | 10 | 40 | 20 | |
The second row (supply from S2) has the largest difference, so we'll select it.
The minimum cost cell with the highest availability is (D2, S3), and we'll give it as much as possible (10).
D1 | D2 | D3 | D4 | S1 | 40 | 10 | 30 | 20 | S2 | 30 | 60 | 20 | 30 | S3 | 0 | 0 | 10 | 40 |
Availability | 20 | 40 | 0 | 80 |
Requirement | 50 | 30 | 40 | 120 |
We'll now update the differences based on the available cells (we only have one remaining).
Supply (units) | Demand (units) | S1 | S2 | S3 | Availability | D1 | 40 | 0 | 30 | 110 | 20 | D2 | 0 | 60 | 0 | 90 | 20 | D3 | 30 | 20 | 0 | 50 | 10 | D4 | 20 | 0 | 10 | 90 | 30 |
Requirement | 50 | 0 | 40 | 120 |
Difference | 10 | 10 | 10 | |
There is only one available row left, so we'll select the first one and provide as much as possible to the minimum cost cell (D1, S2), which is 10.
We'll cross it out and update the availability and demand rows.
D1 | D2 | D3 | D4 | S1 | 30 | 20 | 30 | 20 | S2 | 30 | 50 | 20 | 30 | S3 | 0 | 0 | 10 | 40 |
Availability | 10 | 30 | 0 | 60 |
Requirement | 40 | 0 | 40 | 120 |
The final solution, with a minimum cost of 2050 units, is shown below:
D1 | D2 | D3 | D4 | S1 | 30 | 20 | 30 | 20 | S2 | 0 | 60 | 20 | 30 | S3 | 10 | 0 | 10 | 40 | Total Cost | 1800 | 600 | 650 | 2050 |
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Evaluate each expression using the values given in the table. 1 х f(x) g(x) -3 -2 -4 -3 3 - 1 -2. 0 0 -1 1 ON a. (fog)(1) d.(gof)(0) b. (fog)(-1) e. (gog)(-2) c. (gof)(-1) f. (fof)(-1)
Evaluating each expression using the values given in the table :
a. (f ∘ g)(1) = -2
b. (f ∘ g)(-1) = 3
c. (g ∘ f)(-1) = 0
d. (g ∘ f)(0) = -1
e. (g ∘ g)(-2) = 1
f. (f ∘ f)(-1) = 3
Here is the explanation :
To evaluate each expression, we need to substitute the given values into the functions f(x) and g(x) and perform the indicated composition.
a. (f ∘ g)(1):
First, find g(1) = 0.
Then, substitute g(1) into f: f(g(1)) = f(0) = -2.
b. (f ∘ g)(-1):
First, find g(-1) = 1.
Then, substitute g(-1) into f: f(g(-1)) = f(1) = 3.
c. (g ∘ f)(-1):
First, find f(-1) = 1.
Then, substitute f(-1) into g: g(f(-1)) = g(1) = 0.
d. (g ∘ f)(0):
First, find f(0) = -2.
Then, substitute f(0) into g: g(f(0)) = g(-2) = -1.
e. (g ∘ g)(-2):
First, find g(-2) = -1.
Then, substitute g(-2) into g: g(g(-2)) = g(-1) = 1.
f. (f ∘ f)(-1):
First, find f(-1) = 1.
Then, substitute f(-1) into f: f(f(-1)) = f(1) = 3.
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"As attendance at school drops, so does achievement" is an example of what type of correlation? Negative Positive No correlation
The statement "As attendance at school decreases, achievement also decreases" exemplifies a negative correlation between attendance and achievement.
Correlation pertains to the association or connection between two variables. In this case, the variables are attendance at school and achievement. A negative correlation means that as one variable decreases, the other variable also decreases.
The statement suggests that as attendance at school drops, achievement also decreases. This implies that there is a negative relationship between attendance and achievement. When students attend school less frequently, their academic performance tends to decline.
Negative correlations are characterized by an inverse relationship between variables, where an increase in one variable corresponds to a decrease in the other. In this scenario, the negative correlation indicates that lower attendance is associated with lower achievement levels.
It is important to note that correlation does not imply causation, and there may be other factors influencing both attendance and achievement.
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Find the general solution of y(4) — 4y"" + 2y" - 12y' + 45y = 0
The general solution of the given fourth-order linear homogeneous differential equation is given by y(t) = c₁e^(3t) + c₂e^(5t) + c₃e^(-2t)cos(4t) + c₄e^(-2t)sin(4t), where c₁, c₂, c₃, and c₄ are constants.
To find the general solution of the fourth-order linear homogeneous differential equation y⁽⁴⁾ - 4y″ + 2y″ - 12y′ + 45y = 0, we first solve the characteristic equation to obtain the roots. Based on the nature of the roots, we apply the appropriate methods to find the general solution.
The characteristic equation for the given differential equation is r⁴ - 4r³ + 2r² - 12r + 45 = 0. To solve this equation, we can use various methods such as factoring, synthetic division, or the quadratic formula. By finding the roots of the characteristic equation, we obtain the characteristic roots.
Depending on the nature of the roots, we can classify the solutions into different cases. If all roots are distinct, the general solution is of the form y(x) = c₁e^(r₁x) + c₂e^(r₂x) + c₃e^(r₃x) + c₄e^(r₄x), where c₁, c₂, c₃, and c₄ are constants determined by the initial conditions.
If the roots are repeated, we can include additional terms with higher powers of x in the general solution. For example, if we have a repeated root r with multiplicity m, the general solution includes terms of the form cₙxⁿe^(rx), where n ranges from 0 to m-1.
In some cases, complex roots may appear, leading to solutions involving sine and cosine functions. These complex roots appear in conjugate pairs, and the general solution includes terms of the form c₁e^(αx)cos(βx) + c₂e^(αx)sin(βx), where α and β are real numbers.
By finding the roots of the characteristic equation and applying the appropriate methods based on the nature of the roots, we can determine the general solution of the given fourth-order linear homogeneous differential equation.
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1. A machine produces metal rods used in an automobile suspension system. A random sample of 15 rods is selected, and the diameter is measured. The resulting data (in millimeters) are as follows:
8.24, 8.25, 8.20, 8.23, 8.24, 8.21, 8.26, 8.26, 8.20, 8.25, 8.23, 8.23, 8.19, 8.36, 8.24.
You have to find a 95% two-sided confidence interval on mean rod diameter. What is the upper value of the 95% CI of mean rod diameter? Please report your answer to 3 decimal places.
The upper value of the 95% CI of the mean rod diameter is approximately 8.276 millimeters.
To find the upper value of the 95% confidence interval (CI) of the mean rod diameter, we can use the formula:
Upper CI = sample mean + margin of error
First, we calculate the sample mean. Adding up all the measured diameters and dividing by the sample size gives us:
Sample mean = (8.24 + 8.25 + 8.20 + 8.23 + 8.24 + 8.21 + 8.26 + 8.26 + 8.20 + 8.25 + 8.23 + 8.23 + 8.19 + 8.36 + 8.24) / 15 = 8.2353 (rounded to 4 decimal places)
Next, we need to calculate the margin of error. Since we have a sample size of 15, we can use the t-distribution with 14 degrees of freedom (n - 1) for a 95% confidence level. Consulting the t-distribution table or using statistical software, we find that the critical value for a two-sided 95% CI is approximately 2.145.
The margin of error is then given by:
Margin of error = critical value * (sample standard deviation / √n)
From the given data, the sample standard deviation is approximately 0.0489. Plugging in the values, we have:
Margin of error = 2.145 * (0.0489 / √15) ≈ 0.0407 (rounded to 4 decimal places)
Finally, we calculate the upper CI:
Upper CI = 8.2353 + 0.0407 ≈ 8.276 (rounded to 3 decimal places)
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Thirteen years ago, you deposited $2400 into a superannuation
fund. Eight years ago, you added an additional $1000 to this
account. You earned 8%, compounded annually, for the first five
years, and 5.
The total amount money after thirteen years of savings will be $5030.63
To calculate the amount of money in the account today, we need to calculate the future value of each contribution separately and then add them together.
Let's start by calculating the future value of the initial deposit of $2400 over the first five years at an interest rate of 8% compounded annually.
Using the formula for compound interest:
Future Value = [tex]Principal[/tex] * [tex](1 + Interest Rate)^{Time}[/tex]
Future Value = $2400 * (1 + 0.08)⁽⁵⁾
Future Value = $2400 * (1.08)⁵
Future Value = $2400 * 1.46933
Future Value = $3526.40
So, after five years, the initial deposit will grow to $3526.40.
Now, let's calculate the future value of the additional deposit of $1000 over the last eight years at an interest rate of 5.5% compounded annually.
Future Value = $1000 * (1 + 0.055)⁸
Future Value = $1000 * (1.055)⁸
Future Value = $1000 * 1.50423
Future Value = $1504.23
So, after eight years, the additional deposit will grow to $1504.23.
Now, let's add the two amounts together to find the total amount in the account today:
Total Amount = $3526.40 + $1504.23
Total Amount = $5030.63
So, the total amount money after thirteen years of saving will be $5030.63
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Complete question:
Thirteen years ago, you deposited $2400 into a superannuation fund. Eight years ago, you added an additional $1000 to this account. You earned 8%, compounded annually, for the first five years, and 5.5%, compounded annually, for the last eight years. How much money do you have in your account today?