ok soooo
when you rrly think abt it we all have kicked a pregnant lady

The probability that in the next 18 samples, exactly 2 contain the pollutant is

P(x = 2) =[tex]C(18, 2) (0.1)^2 (0.9)^16[/tex]

= 0.252.

The given is Each sample of water has a 10% chance of containing a particular organic pollutant.

Assume that the samples are independent with regard to the presence of the pollutant.

We need to find the probability that in the next 18 samples, exactly 2 contain the pollutant.

We can use the Binomial distribution formula to solve this problem.

The binomial distribution formula is given by:

P(x) = [tex]C(n, x) px qn-x[/tex]

Where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure.

Since there is a 10% chance of containing a particular organic pollutant, the probability of success is:

p = 0.1

The probability of failure is:

q = 1 - p

= 1 - 0.1

= 0.9

Number of trials (samples) is n = 18

(a) To find P(x≥≥4)

We need to find the probability that 2 or more samples contain the pollutant.

We can do this by subtracting the probability that less than 2 samples contain the pollutant from 1.

P(x ≥ 2) = 1 - P(x < 2)P(x < 2)

= P(x = 0) + P(x = 1)P(x = 0)

= C(18, 0) (0.1)0 (0.9)18

= [tex]0.9^18[/tex]

P(x = 1) = [tex]C(18, 1) (0.1)^1 (0.9)^17[/tex]

= 0.33

P(x < 2) = [tex]0.9^18[/tex]+ 0.33

= 0.335

P(x ≥ 2) = 1 - 0.335

= 0.665

(b) To find P(x≤≤4)

We need to find the probability that 4 or less samples contain the pollutant.

P(x ≤ 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)P(x = 0)

= C(18, 0) (0.1)0 (0.9)18

= [tex]0.9^18[/tex]

P(x = 1) = [tex]C(18, 1) (0.1)^1 (0.9)^17[/tex]

= 0.33

P(x = 2) = [tex]C(18, 2) (0.1)^2 (0.9)^16[/tex]

= 0.053

P(x = 3) = [tex]C(18, 3) (0.1)^3 (0.9)^15[/tex]

= 0.0047

P(x = 4) =[tex]C(18, 4) (0.1)^4 (0.9)^14[/tex]

= 0.0003

P(x ≤ 4) = 0.9^18 + 0.33 + 0.053 + 0.0047 + 0.0003

= 0.288

(d) To find P(3≤≤x≤≤7)

We need to find the probability that between 3 and 7 (inclusive) samples contain the pollutant.

P(3 ≤ x ≤ 7) = P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7)P(x = 3)

= [tex]C(18, 3) (0.1)^3 (0.9)^15[/tex]

= 0.0047

P(x = 4) = C(18, 4) (0.1)^4 (0.9)^14

= 0.0003

P(x = 5) = [tex]C(18, 5) (0.1)^5 (0.9)^13[/tex]

= 0.00001

P(x = 6) = [tex]C(18, 6) (0.1)^6 (0.9)^12[/tex]

= 2.15e-07

P(x = 7) = [tex]C(18, 7) (0.1)^7 (0.9)^11[/tex]

= 4.7e-10

P(3 ≤ x ≤ 7) = 0.0047 + 0.0003 + 0.00001 + 2.15e-07 + 4.7e-10

= 0.00471

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Step-by-step explanation:

D. In 2 _ In

maaf kalo salah

Answer:

Yield per share = 7.68% (Approx.)

Step-by-step explanation:

Given:

Dividend paid = $4.15

Price per dividend = $54

Find:

Yield per share

Computation:

Yield per share = [Dividend paid / Price per dividend]100

Yield per share = [4.15 / 54]100

Yield per share = [0.0768]100

Yield per share = 7.68% (Approx.)

Answer: the answer is B!!!!!!!!!!!!!!

Step-by-step explanation:

The reformulated problem includes the original variables x₁, x₂, x₃, and the additional slack variables s₁, s₂, and s₃. The objective function and constraints remain the same, but the problem has been transformed into an equality form suitable for applying the big-M method or other linear programming techniques.

To reformulate the given linear programming problem using the big-M method, we introduce additional binary variables and big-M constants. The reformulated problem is as follows:

maximize 2x₁ + x₂ + x₃

subject to:

x₂ + x₃ + s₁ = 5

x₁ - x₂ + s₂ = 1

x₁ + x₂ + 3x₃ + s₃ = 25

x₁, x₂, x₃, s₁, s₂, s₃ ≥ 0

where s₁, s₂, and s₃ are slack variables, and we choose a large positive constant M to represent infinity.

Introducing slack variable s₁ for the inequality constraint x₂ + x₃ ≥ 5:

We add s₁ to the left-hand side of the inequality to transform it into an equation: x₂ + x₃ + s₁ = 5.

Introducing slack variable s₂ for the inequality constraint x₁ - x₂ ≥ 1:

We subtract s₂ from both sides of the inequality to transform it into an equation: x₁ - x₂ + s₂ = 1.

Introducing slack variable s₃ for the inequality constraint x₁ + x₂ + 3x₃ ≤ 25: We subtract s₃ from the left-hand side of the inequality to transform it into an equation: x₁ + x₂ + 3x₃ + s₃ = 25.

Since all variables in the original problem are non-negative (x₁, x₂, x₃ ≥ 0), we explicitly state this in the reformulation.

Therefore, the reformulated problem includes the original variables x₁, x₂, x₃, and the additional slack variables s₁, s₂, and s₃. The objective function and constraints remain the same, but the problem has been transformed into an equality form suitable for applying the big-M method or other linear programming techniques.

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Incomplete question:

Reformulate the following LP by the big-M method (only the reformulation is required, no need to solve the LP).

maxₓ 2x₁ + x₂ + x₃

s.t.

x₂+x₃ ≥ 5

x₁ - x₂ ≥ 1

x₁ + x₂+ 3x₃ ≤ 25

x₁, x₂, x₃  ≥0

Answer:

x = 37 mm

Step-by-step explanation:

using the Pythagorean theorem:

m² = 13² - 5² = 169 - 25 = 144

m = √144 = 12 mm

x² = 12² + 35² = 144 + 1225 = 1369

x = √1369 = 37 mm

Answer:

x=37.3

Step-by-step explanation:

Since we're looking for "x" we need to use "the Pythagorean theorem".

13mm will be our "A" and 35mm will be our "B" will "x" being the hypotenuse.

Since x is the hypotenuse, it needs to be a number greater than 13 and 35.

c=√(a² + b²)

Making it c=√(13² + 35²)

Making C=37.336 or 37.3

Therefore making, x=37.3

Answer:

She saved 1.20

Step-by-step explanation:

Purchase Price:

$12

Discount:

(12 x 10)/100 = $1.20

Final Price:

12 - 1.20 = $10.80

Answer:

it's 1,-1

Step-by-step explanation:

hoped I helped. I'm good at these.

When the total number of units produced over the five-year period is 125,000 units, the depreciation will be $1.20 pee unit.

The total number of units produced over the five-year period is 125,000 units. Therefore, the depreciation expense for each year would be calculated as follows:

Depreciation expense = (Cost of asset - Residual value) / Total number of units produced

Depreciation expense = ($150,000 - $12,000) / 125,000 units

Depreciation expense = $1.20 per unit

There are several factors that a company should consider when selecting a depreciation method. These factors include:

The type of asset being depreciated is an important factor to consider because different assets have different useful lives and expected patterns of depreciation. For example, a machine that is expected to be used for five years would be depreciated using a different method than a building that is expected to be used for 30 years.

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Answer:

Step-by-step explanation:

My insight on this quote is that a good piece of art can tell various stories. Maybe there are various characters in the art and each on shows different actions and emotions. Or maybe the art is abstract and the emotions that it triggers in a person can tell a story all on its own. In a way these various angles are the different ways that the art makes us feels that allows us to view it in such a way. Yet these views can change if we look at the art in a different perspective.

Answer:

42

Step-by-step explanation:

Answer:

42 square feet

Step-by-step explanation:

add all of the square feet together, including the sides parallel that don't tell you how many inches. What I mean by parallel is the sides across from it.

Answer:

All are true

Step-by-step explanation:

2) True

3) True

4) True

The student is implicitly using the commutative property and the associative property of addition.

i. (9-1) + 9: The student uses the commutative property, which states that the order of adding numbers does not affect the sum. They rearrange the terms to (9-1) + 9, recognizing that adding 9 after subtracting 1 is the same as adding 9 before subtracting 1.

ii. (9-1) + 9 = 9 + (9-1): Here, the student demonstrates the associative property. The associative property states that the grouping of numbers being added does not affect the sum. They regroup the terms to show that adding (9-1) first and then adding 9 is equivalent to adding 9 first and then subtracting 1.

iii. 9 + (9-1): The student does not explicitly demonstrate a property here. They simply perform the calculation of adding 9 and (9-1), recognizing that 9 minus 1 is 8.

iv. (9+9)-1-18-1: In this step, the student performs the subtraction calculations but does not demonstrate any particular property.

v. 18-1: Finally, the student calculates the subtraction and arrives at the answer 17.

So, the student implicitly uses the commutative property and the associative property of addition in their mental calculation.

Answer:

15 feet

Step-by-step explanation:

The distance from the foot of the ladder to the person is 6 feet. The person is also 6 feet. Therefore that mini triangle's length and width are both 6 feet.

The distance from the foot of the ladder to the wall is 15 feet. This is now a simple ratio. The length of the mini triangle is 6 to length of the larger triangle, 15. The height ratio is the mini triangle is 6 to the bigger triangle, x.

The height of the bigger triangle is the height of the wall. (Since the two are similar, then we can automatically deduce the height is 15.) But if you wanted to work out the ratio above ^^ you would multiply the 6 diagnol to the 15 and then divide by the last 6 to get 15.

Answer:

the graph has two vertical asymptotes, line q intersects the line at -8 and is the more important one.

Step-by-step explanation:

This is visible based off of the picture.

Answer:

P- arenthesis

E- xpontent

M- ultiplication

D- ivision

A- ddition

S- ubtraction

Answer:

Step-by-step explanation:

p=parentheses

e=exponents

m=multiplication

d=division

a=addition

s=subtraction

Answer:

Monday - 7 hr 30 mins

Wednesday - 9 hr 30 mins

Friday - 7 hr

Saturay - 5 hr 30 mins

---------------------------------------------------------------

If you earn $12.40 per hour, first you must find how many hours you worked total - in this case, 29 hr and 30 mins.

So you would be making $359.60 per week.

For x = 9a - 7, where a is an integer, the value of x mod 3 is determined using the remainder when x is divided by 3, and it is equal to 2.

To find the value of x mod 3, we need to determine the remainder when x is divided by 3.

The expression x = 9a - 7, where a is an integer, allows us to calculate x mod 3.

To find x mod 3, we substitute the given expression x = 9a - 7 into the equation:

x mod 3 = (9a - 7) mod 3.

To simplify the expression, we can use the properties of modular arithmetic.

First, we can rewrite 9a as (3 × 3a) and distribute the modulus operator:

(9a - 7) mod 3 = ((3 × 3a) - 7) mod 3.

Next, we can separate the modulus operation:

((3 × 3a) - 7) mod 3 = ((3 × 3a) mod 3 - 7 mod 3) mod 3.

Since 3 is congruent to 0 mod 3, the first term becomes 0:

((3 × 3a) mod 3 - 7 mod 3) mod 3 = (0 - 7 mod 3) mod 3.

The second term, 7 mod 3, can be simplified as the remainder when 7 is divided by 3, which is 1:

(0 - 7 mod 3) mod 3 = (0 - 1) mod 3 = -1 mod 3.

Finally, we can find the remainder when -1 is divided by 3, which is 2:

-1 mod 3 = 2.

Therefore, x mod 3 is equal to 2.

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After 100 years, the long-term distribution of population in the city and suburbs of New York state can be calculated based on the given migration rates. The population in the city and suburbs will stabilize at approximately 3.96 million and 16.04 million, respectively. The population distribution can be visualized using a graph that shows the population of the city and suburbs over the 100-year period.

To calculate the long-term population distribution, we can use the concept of equilibrium. Let C represent the population in the city and S represent the population in the suburbs. The equilibrium equations can be written as follows:

C = C - 0.035C + 0.017S

S = S + 0.035C - 0.017S

Simplifying these equations, we have:

C = 0.965C + 0.017S

S = 0.035C + 0.983S

Solving these equations simultaneously, we find that C stabilizes at approximately 3.96 million and S stabilizes at approximately 16.04 million.

To plot the population of the city and suburbs over the 100-year period, you can use the following MATLAB code:

Copy code

years = 0:100;

C = zeros(1, 101);

S = zeros(1, 101);

C(1) = 8.4;

S(1) = 20 - C(1);

for i = 2:101

   C(i) = 0.965*C(i-1) + 0.017*S(i-1);

   S(i) = 0.035*C(i-1) + 0.983*S(i-1);

end

plot(years, C, 'b', 'LineWidth', 2);

hold on;

plot(years, S, 'r', 'LineWidth', 2);

xlabel('Years');

ylabel('Population');

legend('City', 'Suburbs');

title('Population of City and Suburbs Over 100 Years');

This MATLAB code calculates and plots the population of the city (in blue) and suburbs (in red) over the 100-year period.

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The probability that the selected individual has children in public schools AND favors the school tax is 0.32

The probability that the selected individual favors the school tax AND has children in public schools is 0.32.

The probability that the selected individual favors the school tax AND does NOT have children in public schools is 0.2.

The probability that the selected individual favors the school tax AND is over 56 years old is 0.15.

The probability that the selected individual favors the school tax AND is 18-25 years old is 0.45.

Based on the given information, the probability of event F (the selected individual favors the school tax) is 0.54, as 54% of the respondents are in favor of the tax. The probability of event C (the selected individual has children in public schools) is 0.59, as the approval rating rises to 59% for those with children in public schools. The probability of event O (the selected individual is over 56 years old) is 0.37, as only 37% of those over age 56 said they would vote for the tax. The probability of event Y (the selected individual is 18-25 years old) is 0.71, as 71% of 18- to 25-year-olds said they would vote for the tax.

Using these probabilities, we can estimate the values of the following probabilities:

(1) P(CF) is the probability that the selected individual has children in public schools AND favors the school tax. Based on the given information, we can multiply the probabilities of events C and F: P(CF) = 0.59 * 0.54 = 0.318, or approximately 0.32.

(ii) P(FIC) is the probability that the selected individual favors the school tax AND has children in public schools. This is the same as P(CF), so P(FIC) = 0.32.

(iii) P(FIN) is the probability that the selected individual favors the school tax AND does NOT have children in public schools. To calculate this, we can use the fact that the approval rating falls to 44% for those with no children in public schools. So, P(FIN) = 0.44 * (1 - 0.59) = 0.18, or approximately 0.2.

(iv) P(FTO) is the probability that the selected individual favors the school tax AND is over 56 years old. To calculate this, we can use the fact that the approval rating for those over 56 years old is only 37%. So, P(FTO) = 0.37 * (1 - 0.59) = 0.1523, or approximately 0.15.

(v) P(FY) is the probability that the selected individual favors the school tax AND is 18-25 years old. To calculate this, we can use the fact that the approval rating for those 18-25 years old is 71%. So, P(FY) = 0.71 * (1 - 0.37) = 0.4477, or approximately 0.45.

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Answer:

1/5

Step-by-step explanation:

i searched it up yikessssssss

The value of the given term is 1296.

A fraction consisting of a quotient and remainder is a mixed fraction. we can convert the mixed fraction to improper fraction by first dividing the numerator by denominator and then taking the quotient as whole number and remainder as the numerator of proper fraction keeping the denominator same.

We are given the number as [tex]\dfrac{1}{6} ^{-4}[/tex]

Here we can see that the power is negative so we can just reciprocal the term.

Then we get;

6^4

this means 6 x6x6x6 = 1296

The correct value is 1296.

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Yes, there is a uniformly most powerful (UMP) test at level a for testing for the given density function.

The test is based on the likelihood ratio, where the critical region is (θ_o, ∞) and the test statistic is (nθ_o^3)/Y(n), where Y(n) is the largest observation in the sample.

To obtain the UMP test at level a for testing H_0: θ ≤ θ_o vs. H_1: θ > θ_o, we need to find the likelihood ratio test with the largest power for all possible alternatives. The likelihood ratio test is constructed as the ratio of the likelihood function under H_0 to the likelihood function under H_1. By algebraic manipulation, we obtain the likelihood ratio test statistic as (nθ_o^3)/Y(n), where Y(n) is the largest observation in the sample.

Under H_0, this test statistic has a chi-squared distribution with one degree of freedom. Therefore, the critical region for rejecting H_0 at level a is the right tail of the chi-squared distribution with one degree of freedom, which is (θ_o, ∞).

This test is UMP because it has the highest power for all possible alternatives. This is because the distribution of Y(n) is stochastically increasing in θ, which means that for a given sample size n, the probability of obtaining an observation larger than a threshold value increases as θ increases.

Therefore, the likelihood ratio test statistic decreases as θ increases, which means that the rejection region (θ_o, ∞) has the highest power for all possible alternatives. Hence, the test based on the likelihood ratio is the UMP test at level a for testing H_0: θ ≤ θ_o vs. H_1: θ > θ_o for the given density function

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Answer:  Complementary     x=15

Step-by-step explanation:

Complementary angles add up to 90°, supplementary angles add up to 180°.

We know they add up to 90 so...

3x+45=90

3x=45

x=15

Answer:

25

Step-by-step explanation:

First, plug -5 in for a, -(-5)^2. We treat the negative on the outside of the paranthese as a -1 so we do -1 times -5 and we get 5. Then we square 5 and get 25.

Question:

A population consists  1, 2, 4, 5, 8. Draw all possible samples of size 2  without replacement from this population.

Verify that the sample mean is an unbiased estimate of the population mean.  

Answer:

[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]

[tex]\hat p = \frac{3}{5}[/tex] --- proportion of evens

The sample mean is an unbiased estimate of the population mean.

Step-by-step explanation:

Given

[tex]Numbers: 1, 2, 4, 5, 8[/tex]

Solving (a): All possible samples of 2 (W.O.R)

W.O.R means without replacement

So, we have:

[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]

Solving (b): The sampling distribution of the proportion of even numbers

This is calculated as:

[tex]\hat p = \frac{n(Even)}{Total}[/tex]

The even samples are:

[tex]Even = \{2,4,8\}[/tex]

[tex]n(Even) = 3[/tex]

So, we have:

[tex]\hat p = \frac{3}{5}[/tex]

Solving (c): To verify

[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]

Calculate the mean of each samples

[tex]Sample\ means = \{1.5,2.5,3,4.5,3,3.5,5,4.5,6,6.5\}[/tex]

Calculate the mean of the sample means

[tex]\bar x = \frac{1.5 + 2.5 +3 + 4.5 + 4 + 3.5 + 5 + 4.5 + 6 + 6.5}{10}[/tex]

[tex]\bar x = \frac{40}{10}[/tex]

[tex]\bar x = 4[/tex]

Calculate the population mean:

[tex]Numbers: 1, 2, 4, 5, 8[/tex]

[tex]\mu = \frac{1 +2+4+5+8}{5}[/tex]

[tex]\mu = \frac{20}{5}[/tex]

[tex]\mu = 4[/tex]

[tex]\bar x = \mu = 4[/tex]

This implies that [tex]\bar x[/tex] is an unbiased estimate of the [tex]\mu[/tex]

Answer:

325 feet below sea level i believe

Step-by-step explanation:

Answer:

The diameter would be 305 m

Step-by-step explanation:

Answer:

305 assuming your using 3.14 for pi

Step-by-step explanation:

Probability of ultimate extinction: To get the probability of ultimate extinction in a branching process, we need to calculate the value of o(q) = q where q is the probability of ultimate extinction.

Now, o(s) = (1/6) + (5/6)s²

Therefore, for the probability of ultimate extinction,

o(q) = q = (1/6) + (5/6)q²=> 6q² - 5q + 1 = 0

On solving the quadratic equation,

we get q = 1 or q = 1/6.

Thus, the probability of ultimate extinction is 1/6. Mean time to extinction:

To calculate the mean time to extinction, we can use the formula, E(T) = (1/o'(1))

where o'(1) is the first derivative of the generating function o(s) evaluated at s = 1.

Now, o(s) = (5/6) + (1/6)s=> o'(s) = (1/6)

On substituting s = 1, o'(1) = (1/6).

Thus, E(T) = (1/o'(1))= (1/ (1/6))= 6

The mean time to extinction is 6.

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