The minimum sample size required is 94 to obtain a margin of error of 0.04236.
To find the minimum sample size required to obtain a margin of error of 0.04236.
we need to use the formula for sample size determination for proportions:
n = (Z² × p × (1 - p)) / E²
Where:
n = required sample size
Z = Z-score corresponding to the desired level of confidence (we'll assume a 95% confidence level, so Z ≈ 1.96)
p = estimated proportion (from the available data)
E = margin of error
Given:
Total adults in the town = 88
Adults with health insurance = 69
First, we calculate the estimated proportion (p) of adults with health insurance:
p = (69 / 88)
Next, we substitute the values into the formula and solve for n:
n = (1.96² × (69/88) × (1 - 69/88)) / (0.04236²)
n = 93.966
The minimum sample size required is 94 to obtain a margin of error of 0.04236 (or 4.236%).
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A sphere has a diameter of 4(x+3) centimeters and a surface area of
784 square centimeters. Find the value of x.
Answer:
78 square centimeters
Step-by-step explanation:
Please answer correctly! I will mark you Brainliest!
Answer:
V=385 cubic units
Step-by-step explanation:
The volume of a rectangular prism is given by the formula [tex]V=lwh[/tex], where l is the length, w is the width, and h is the height. Our dimensions are 5, 7, and 11. So, we have to multiply them to find the volume.
5 × 7 × 11 = 385
Thus, the volume of the rectangular prism is 385 cubic units.
Gerald and Wheatly, Applied Numerical Analysis ▶6. If e¹.3 is approximated by Lagrangian interpolation from the values for eº = 1, el = 2.7183, and e² = 7.3891, what are the minimum and maximum estimates for the error? Compare to the actual error.
Lagrangian interpolation is used to approximate the value of e¹.3 using three known values: eº = 1, el = 2.7183, and e² = 7.3891. We can find the minimum and maximum estimates for the error.
To approximate e¹.3 using Lagrangian interpolation, we construct a polynomial that passes through the three given points: (0, 1), (1, 2.7183), and (2, 7.3891). We can then evaluate this polynomial at x = 1.3 to estimate the value of e¹.3.
Using Lagrangian interpolation, the polynomial P(x) is given by:
P(x) = 1 * L₀(x) + 2.7183 * L₁(x) + 7.3891 * L₂(x),
where L₀(x), L₁(x), and L₂(x) are the Lagrange basis polynomials associated with the three data points.
To find the minimum and maximum estimates for the error, we need to determine the upper bound for the error term in the Lagrangian interpolation formula. The error term is given by:
E(x) = f(x) - P(x),
where f(x) is the actual function we are approximating (in this case, e^x).
To find the upper bound for the error, we can use the maximum value of the absolute value of the n+1st derivative of f(x) in the interval containing the data points.
By calculating the upper bound for the error, we can compare it to the actual error by evaluating the actual function e¹.3 and subtracting the approximation P(1.3) obtained from Lagrangian interpolation.
By analyzing the error estimates and comparing them to the actual error, we can assess the accuracy of the Lagrangian interpolation approximation in this particular case.
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what would be the distance between (-35,20) and (15,20).
Answer:
Well you just have to follow the formula to find the distance between 2 linear plots.
d= √(x2−x1)^2+(y2−y1)^2
Step-by-step explanation:
If you follow it the answer would be
D=50
Solve the following differential equation by using Laplace transform method. y" +2y' +y = cos2t where y(0)=1 y'(O)=1.
The solution to the given differential equation with the initial conditions y(0) = 1 and y'(0) = 1 is:
[tex]y(t) = -e^{-t}/10 + (11/10)t*e^{-t} + (1/10)cos(2t) + (3/10)sin(2t)[/tex]
To solve the given differential equation using Laplace transform, we will apply the Laplace transform to both sides of the equation and then solve for the transformed variable.
Let's denote the Laplace transform of y(t) as Y(s).
Taking the Laplace transform of both sides of the differential equation, we get:
[tex]s^2Y(s) + 2sY(s) + Y(s) = (s^2 + 2s + 1)/(s^2 + 4)[/tex]
Now, let's solve for Y(s):
[tex]Y(s)(s^2 + 2s + 1) = (s^2 + 2s + 1)/(s^2 + 4)\\Y(s) = (s^2 + 2s + 1)/(s^2 + 4)(s^2 + 2s + 1)[/tex]
Factoring the denominator:
[tex]Y(s) = (s^2 + 2s + 1)/((s + 1)^2(s^2 + 4))[/tex]
Now, we need to decompose the fraction into partial fractions. Let's express the numerator in terms of A, B, C, and D:
[tex]s^2 + 2s + 1 = A/(s + 1) + B/(s + 1)^2 + (Cs + D)/(s^2 + 4)[/tex]
To find the values of A, B, C, and D, we can equate the numerators:
[tex]s^2 + 2s + 1 = A(s + 1)(s^2 + 4) + B(s^2 + 4) + (Cs + D)(s + 1)^2[/tex]
Expanding and equating coefficients:
[tex]s^2 + 2s + 1 = A(s^3 + 5s^2 + 4s) + B(s^2 + 4) + (C(s^2 + 2s + 1) + D(s + 1)^2)[/tex]
Simplifying:
[tex]s^2 + 2s + 1 = (A + C)s^3 + (5A + C + D)s^2 + (4A + 2C + D)s + (4A + D)[/tex]
Equating coefficients:
A + C = 0 (coefficient of [tex]s^3[/tex])
5A + C + D = 1 (coefficient of [tex]s^2)[/tex]
4A + 2C + D = 2 (coefficient of s)
4A + D = 1 (constant term)
Solving these equations simultaneously, we find A = -1/10, B = 11/10, C = 1/10, and D = 3/10.
Now, substituting these values back into Y(s):
[tex]Y(s) = (-1/10)/(s + 1) + (11/10)/(s + 1)^2 + (1/10)(s + 3)/(s^2 + 4) + (3/10)/(s^2 + 4)[/tex]
To find y(t), we need to take the inverse Laplace transform of Y(s). Fortunately, we can use a Laplace transform table to find the inverse Laplace transform of each term.
The inverse Laplace transform of (-1/10)/(s + 1) is [tex]-e^{-t}/10.[/tex]
The inverse Laplace transform of (11/10)/(s + 1)² is (11/10)t*[tex]e^{-t}.[/tex]
The inverse Laplace transform of (1/10)(s + 3)/(s² + 4) is (1/10)cos(2t).
The inverse Laplace transform of (3/10)/(s² + 4) is (3/10)sin(2t).
Combining these results, the solution y(t) is:
[tex]y(t) = -e^{-t}/10 + (11/10)t*e^{-t} + (1/10)cos(2t) + (3/10)sin(2t)[/tex]
Therefore, the solution to the given differential equation with the initial conditions y(0) = 1 and y'(0) = 1 is:
[tex]y(t) = -e^{-t}/10 + (11/10)t*e^{-t} + (1/10)cos(2t) + (3/10)sin(2t)[/tex]
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Use the Laplace transform to solve the initial-value problem x" + 4 = f(t), x(0)=0, x'(0) = 0, if t < 5 f(t) = t25. 3 sin(t-5) if t > 5.
By applying the initial conditions and inverse Laplace transforming, we can obtain the solution x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5), where u(t) is the unit step function. Therefore, the solution to the initial-value problem is x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5)
Taking the Laplace transform of the given differential equation x" + 4 = f(t), we obtain the algebraic equation in the Laplace domain: s^2X(s) + 4sX(s) + 4 = F(s), where X(s) is the Laplace transform of x(t) and F(s) is the Laplace transform of f(t).
Next, applying the initial conditions x(0) = 0 and x'(0) = 0, we get X(0) = 0 and sX(0) = 0. Substituting these initial conditions into the Laplace domain equation, we have s^2X(s) + 4sX(s) + 4 = F(s), with X(0) = 0 and sX(0) = 0.
Now, let's consider the Laplace transform of f(t) using the given piecewise function. For [tex]t < 5, f(t) = t^2/5, and for t > 5, f(t) = 3sin(t-5).[/tex]Taking the Laplace transform of f(t) in each interval, we have [tex]F(s) = (1/s^3) + (3/s^2) for t < 5 and F(s) = (3/s^2) * (1/(s^2+1)) for t > 5.[/tex]
Substituting these Laplace transforms into the equation[tex]s^2X(s) + 4sX(s) +[/tex]4 = F(s), we can solve for X(s). Simplifying, we obtain [tex]X(s) = (1/s^3) + (3/s^2) / (s^2 + 4s + 4) + (3/s^2) * (1/(s^2+1)).[/tex]
To find the inverse Laplace transform of X(s), we can split it into partial fractions and apply the inverse Laplace transform formula. The solution is x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5), where u(t) is the unit step function.
Therefore, the solution to the initial-value problem is x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5), where u(t) is the unit step function that ensures the piecewise function is activated at t = 5.
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The change in water level of a lake is modeled by a polynomial function, W(x). Describe how to find the x-intercepts of W(x) and how to construct a rough graph of W(x) so that the Parks Department can predict when there will be no change in the water level. You may create a sample polynomial of degree 3 or higher to use in your explanations.
(Dividing and Solving Polynomials)
First. Finding the x-intercepts of
Let be the change in water level. So to find the x-intercepts of this function we can use The Rational Zero Test that states:
To find the zeros of the polynomial:
We use the Trial-and-Error Method which states that a factor of the constant term:
can be a zero of a polynomial (the x-intercepts).
So let's use an example: Suppose you have the following polynomial:
where the constant term is . The possible zeros are the factors of this term, that is:
.
Thus:
From the foregoing, we can affirm that are zeros of the polynomial.
Second. Construction a rough graph of
Given that this is a polynomial, then the function is continuous. To graph it we set the roots on the coordinate system. We take the interval:
and compute where is a real number between -2 and -1. If , the curve start rising, if not, the curve start falling. For instance:
Therefore the curve start falling and it goes up and down until and from this point it rises without a bound as shown in the figure below
Ariana orders 4 large pizzas and 1 order of breadsticks The total for her order is $34.46. Emily orders 2 large pizzas and 1 order of breadsticks. $18.48 is the total for her order. Determine the order for 1 large pizza and 1 order of breadsticks?
use the limit comparison test to determine whether the series converges. k^2 2/k^3-7
The series ∑[(k² + 2)/(k³ - 7)] diverges using the limit comparison test and the series ∑(1/k) is a well-known harmonic series.
To determine the convergence of the series ∑[(k² + 2)/(k³ - 7)], we can use the limit comparison test. Let's compare it with the series ∑(1/k).
First, we need to find the limit of the ratio of the terms of the two series as k approaches infinity:
lim(k→∞) [(k² + 2)/(k³ - 7)] / (1/k)
Simplifying the expression, we get:
lim(k→∞) [(k² + 2)/(k³ - 7)] × (k/1)
Taking the limit as k approaches infinity, we have:
lim(k→∞) [(k² + 2)/(k³ - 7)] × (k/1) = 1
Since the limit is a finite positive value (1), we can conclude that the given series ∑[(k² + 2)/(k³ - 7)] and the series ∑(1/k) have the same convergence behavior.
The series ∑(1/k) is a well-known harmonic series, which diverges. Therefore, by the limit comparison test, the given series ∑[(k² + 2)/(k³ - 7)] also diverges.
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The question is -
Use the limit comparison test to determine whether the series converges. k² + 2/ k³ - 7
Consider a population consisting of the following five values, which represent the number of video downloads during the academic year for each of five housemates. 9 15 18 11 12 (a) Compute the mean of this population. H = 13 (b) Select a random sample of size 2 by writing the five numbers in this population on slips of paper, mixing them, and then selecting two. Calculate the mean for your sample. (c) Repeatedly select random samples of size 2, and calculate the x value for each sample until you have the values for 25 samples. Describe your results. This answer has not been graded yet. (d) Construct a density histogram using the 25 x values. Are most of the values near the population mean? Do the values differ a lot from sample to sample, or do they tend to be similar?
(a) Mean = (9 + 15 + 18 + 11 + 12) / 5 = 65 / 5 = 13
(b) Mean of the sample = (9 + 15) / 2 = 24 / 2 = 12
(c) The means of the samples vary, but they tend to be close to the population mean of 13.
(d) Since the problem does not provide the values for the 25 x values, we cannot construct a density histogram or determine if most of the values are near the population mean.
(a) The mean of the population is calculated by summing all the values and dividing by the total number of values:
Mean = (9 + 15 + 18 + 11 + 12) / 5 = 65 / 5 = 13
(b) To calculate the mean for a random sample of size 2, we randomly select two values from the population and calculate their mean:
Random sample: 9, 15
Mean of the sample = (9 + 15) / 2 = 24 / 2 = 12
(c) Repeatedly selecting random samples of size 2 and calculating the mean for each sample:
Here are the means for 25 random samples of size 2 (selected without replacement):
Sample 1: 9, 18 -> Mean = (9 + 18) / 2 = 27 / 2 = 13.5
Sample 2: 11, 9 -> Mean = (11 + 9) / 2 = 20 / 2 = 10
Sample 3: 15, 12 -> Mean = (15 + 12) / 2 = 27 / 2 = 13.5
...
Sample 25: 12, 15 -> Mean = (12 + 15) / 2 = 27 / 2 = 13.5
The means of the samples vary, but they tend to be close to the population mean of 13.
(d) Constructing a density histogram using the 25 x values:
Since the problem does not provide the values for the 25 x values, we cannot construct a density histogram or determine if most of the values are near the population mean.
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What is the Quartile 1 for the Box & Whisker Plot below?
PLSS HELP
Answer:
17
Step-by-step explanation:
The lower quartile Q₁ is positioned at the left side of the box.
The value at the left is Q₁ is 17
I need help . Pretend they are labeled a-e
Answer:
a definitely, and maybe e, im not sure
Step-by-step explanation:
let me know if it was right
Eat your spinach: Six measurements were made of the mineral content (in percent) of spinach, with the following results. It is reasonable to assume that the population is approximately normal. 19.1 20.8 20.8 21.4 20.5 19.7 a. Construct a 95% confidence interval for the mean mineral content. b. Based on the confidence interval, is it reasonable to believe that the mean mineral content of spinach may be greater than 21%? Explain.
a. The 95% confidence interval for the mean mineral content of spinach can be calculated using the sample data. The formula for the confidence interval is:
Confidence interval = (sample mean) ± (critical value) * (standard deviation / sqrt(sample size))
Using the given data, the sample mean is 20.63 and the standard deviation is 0.676. The critical value for a 95% confidence level can be obtained from the t-distribution table for a sample size of 6 (n-1 degrees of freedom). Calculating the confidence interval using these values gives a range of approximately 19.97% to 21.29%.
b. Based on the 95% confidence interval, the range of the mean mineral content of spinach is between 19.97% and 21.29%. Since this range includes the value of 21%, it is reasonable to believe that the mean mineral content of spinach may be greater than 21%. However, we cannot be certain as the range also includes values below 21%. A larger sample size or narrower confidence interval would provide more precise information about the true mean.
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HELP HELP PLEASE WRING ANSWERS OR LINKS WILL BE REPORTED
Answer:
$0.24 per orange
Step-by-step explanation:
$3.84 / 16 = 0.24
Solve the given system of differential equations by systematic elimination.
(D − 1)x+ (D² + 1)y = 1
(D² − 1)x+ (D + 1)y = 2
(x(t), y(t)) = (e^-t/2 [-5/3cos(√47/2)t - 125/3 sin(√47/2)t]+ 20/3 cos (3t) + 20/3 sin (3t)
Given system of differential equation is(D − 1)x+ (D² + 1)y = 1 ...
(i)(D² − 1)x+ (D + 1)y = 2 ...(ii)By using systematic elimination method, we have(D²+1)(D²−1)x+(D+1)(D−1)y=D²+1×1-(D+1)×1=0Simplifying the above equation, we get(D⁴-1)x=-(D-1)y...(iii)Applying D on both sides of (iii), we get D(D⁴-1)x=-(D-1 )DyD⁵x- Dx=-(Dy-y)or D⁵x+Dy=y ... (iv)Now applying D on (i), we get(D−1)Dx+(D²+1)Dy=0or D(D²+1)y=(1-D)x ...(v)Now applying D on (ii), we get(D²−1)Dx+(D+1)Dy=0or D(D+1)x=(1+D)y ...(vi)Now, substituting the value of x and y from equations (v) and (vi) in equation (iv), we getD⁵x+(1+D)Dx=(1-D)Dy D⁵x+(1+D)Dx=-(1-D)x ...(vii) Simplifying the above equation, we getD⁶x+2D⁴x+D²x+x=0or D²(D⁴+1)x+D²x=-x ...(viii)or D²(D⁴+2)x=-xor D⁴x+2x=-xor D⁴x=-3xNow using D on both sides, we get D⁵x=-3Dxor D⁶x=-3D²x
Now, substituting the value of D²x from equation (iii) in equation (i), we get(D-1)x+(D²+1)y=1 ...(i)⇒ (D-1)x+y=1 ...(ix)Now, substituting the value of D²x from equation (iii) in equation (ii), we get(D²-1)x+(D+1)y=2 ...(ii)⇒ -(D+1)x+y=0or (D+1)x-y=0 ...(x)From equation (ix) and (x), we have2x=1or x=1/2Now, substituting the value of x in equation (ix), we have D(1/2)+y=1or y=1-1/2=1/2Thus, the solution of the given system of differential equation is(x(t), y(t))=(e^(-t/2))[(-5/3)cos((sqrt(47)/2)t)-(125/3)sin((sqrt(47)/2)t)]+(20/3)cos(3t)+(20/3)sin(3t), (1/2)
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Arletta built a cardboard ramp
for her little brother’s toy cars.
Find the volume of this ramp.
Answer:
Volume = 525 in³
Step-by-step explanation:
Volume = 25 x 7 x 6 x 0.5 = 525 in³
(f+g)(2)=
(f-g)(2)=
(fg)(2)=
Answer:
(f + g)(2) = 16
(f - g)(2) = 0
(fg)(2) = 64
Step-by-step explanation:
f(2) = 8 g(2) = 8
(f + g)(2) = f(2) + g(2) = 8 + 8 = 16
(f - g)(2) = f(2) - g(2) = 8 - 8 = 0
(fg)(2) = f(2)g(2) = 8(8) = 64
A cube with an edge of length s has a volume of 27 units.
What is the length of s?
Answer:
s = 3
Step-by-step explanation:
The volume formula for a cube is V = s^3, where “s” is the edge length. Since we know the volume and need to find “s,” we just do the inverse operation for an exponent, which is a radical. The cubed root of 27 is 3, so there’s your answer! Hope this is helpful & accurate. Best wishes.
hurry !!!! I need help
Answer:
x = 66
Step-by-step explanation:
m<5 + m<6 = 180
2x + 48 = 180
2x = 132
x = 66
AB
Round your answer to the nearest hundredth.
А
50°
6
12
B
Answer:
h = 7.832
Step-by-step explanation:
This is a right angled triangle so, taking 50 as reference angle,
hypotenuse = ?
perpendicular = 6
The ratio for p and h is given by
Sin 50 = p/h
Sin 50 = 6 /h
h = 6 / Sin 50
h = 7.832
The Taylor polynomial P, = (-10 * 9 about x = 0 is used to approximate the value of the function f at x=1 Find the value that verifies 5p (1)-(1)-500 n=1 n! Pa 1 384 1 384 0 ਤਕ OP PA 1 6144 PA 6144
The value that verifies 5p (1)-(1)-500 is -124.04
To approximate the value of the function f at x = 1 using the Taylor polynomial Pₙ = (-10)^n/ n! about x = 0, we need to find the value of P₅(1).
First, let's compute the derivatives of f(x) = e^x up to the fifth derivative:
f'(x) = e^x
f''(x) = e^x
f'''(x) = e^x
f''''(x) = e^x
f⁽⁵⁾(x) = e^x
Now, let's evaluate these derivatives at x = 0:
f(0) = e^0 = 1
f'(0) = e^0 = 1
f''(0) = e^0 = 1
f'''(0) = e^0 = 1
f''''(0) = e^0 = 1
f⁽⁵⁾(0) = e^0 = 1
Using these values, we can compute the Taylor polynomial P₅(x):
P₅(x) = f(0) + f'(0)(x - 0) + f''(0)(x - 0)²/2! + f'''(0)(x - 0)³/3! + f''''(0)(x - 0)⁴/4! + f⁽⁵⁾(0)(x - 0)⁵/5!
P₅(x) = 1 + 1x + 1x²/2! + 1x³/3! + 1x⁴/4! + 1x⁵/5!
Now, let's evaluate P₅(1):
P₅(1) = 1 + 1(1) + 1(1)²/2! + 1(1)³/3! + 1(1)⁴/4! + 1(1)⁵/5!
P₅(1) = 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120
P₅(1) = 227/120
Therefore, the value that verifies 5P₅(1) - (1) - 500 is:
5P₅(1) - (1) - 500 = 5 * (227/120) - 1 - 500
= 1135/120 - 1 - 500
= 1135/120 - 120/120 - 60000/120
= (1135 - 120 - 60000)/120
= -59485/120
= -124.04
So, the value that verifies 5P₅(1) - (1) - 500 is approximately -124.04.
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Can someone plz help me with the answer
Step-by-step explanation:
[tex] \frac{12 \times 8}{2} + {( \frac{8}{2}) }^{2} \pi = \\ = 48 + 16\pi[/tex]
98.24
can someone please help me answer these?
Answer: w+x=90
z=y+90
Step-by-step explanation: because the sum of the inner angles of triangle is 180. given that one angle is 90 degrees the sun of x and w has to be 180-90=90
for the second part, you have to use the triangle exterior angle theorem: an exterior angle of a triangle is equal to the sum of the opposite interior angles. The opposite interior angles in this case is 90 degrees and y
Assume that the playbook contains 16 passing plays and 12 running plays. The coach randomly selects 8 plays from the playbook. What is the probability that the coach selects at least 3 passing plays and at least 2 running plays?
The probability that the coach selects at least 3 passing plays and at least 2 running plays out of 8 plays from the playbook is approximately 0.4914 or 49.14%. This means there is a 49.14% chance of the coach choosing a combination that meets the given criteria.
To calculate the probability of the coach selecting at least 3 passing plays and at least 2 running plays out of 8 plays, we need to consider different combinations that satisfy these conditions.
1: Determine the total number of possible combinations of 8 plays from a playbook of 28 plays (16 passing plays + 12 running plays).
Total Combinations = C(28, 8) = 28! / (8! * (28-8)!) = 3,395,685
2: Calculate the number of combinations that have at least 3 passing plays and at least 2 running plays.
First, we calculate the number of combinations with exactly 3 passing plays and 2 running plays:
Number of Combinations with 3 passing and 2 running = C(16, 3) * C(12, 2) = (16! / (3! * (16-3)!) * (12! / (2! * (12-2)!) = 560 * 66 = 36,960
Next, we calculate the number of combinations with exactly 4 passing plays and 2 running plays:
Number of Combinations with 4 passing and 2 running = C(16, 4) * C(12, 2) = (16! / (4! * (16-4)!) * (12! / (2! * (12-2)!) = 1,820 * 66 = 120,120
Finally, we calculate the number of combinations with 5 passing plays and at least 2 running plays:
Number of Combinations with 5 passing and 2 or more running = C(16, 5) * (C(12, 2) + C(12, 3) + C(12, 4) + C(12, 5) + C(12, 6) + C(12, 7) + C(12, 8)) = (16! / (5! * (16-5)!) * (C(12, 2) + C(12, 3) + C(12, 4) + C(12, 5) + C(12, 6) + C(12, 7) + C(12, 8)) = 4368 * (66 + 220 + 495 + 792 + 924 + 792 + 495) = 4368 * 3786 = 16,530,048
Total Number of Combinations with at least 3 passing and 2 running plays = Number of Combinations with 3 passing and 2 running + Number of Combinations with 4 passing and 2 running + Number of Combinations with 5 passing and 2 or more running = 36,960 + 120,120 + 16,530,048 = 16,687,128
3: Calculate the probability.
Probability = (Number of Combinations with at least 3 passing and 2 running plays) / (Total Combinations) = 16,687,128 / 3,395,685 ≈ 0.4914
Therefore, the probability that the coach selects at least 3 passing plays and at least 2 running plays out of 8 plays is approximately 0.4914 or 49.14%.
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1. Find the equation of the parabola satisfying the given conditions.
Focus: (3,6); Directrix: x=−1
A. (x−1)2=8(y−6)
B. (y−6)2=8(x−1)
C. (x−1)2=−8(y−6)
D. (y−6)2=−8(x−1)
2. Find the equation of the parabola satisfying the given conditions.
Focus: (−6,3); Directrix: y=1
A. (y−2)2=4(x+6)
B. (x+6)2=4(y−2)
C. (x+6)2=−4(y−2)
D. (y−2)2=−4(x+6)
3. Find the equation of an ellipse that has foci at (−1,0) and (4,0), where the sum of the distances between each point on the ellipse and the two foci is 9.
A. (x+1)2+y2−−−−−−−−−−−√+(x−4)2+y2−−−−−−−−−−−√=9
B. (x−1)2+y2−−−−−−−−−−−√+(x+4)2+y2−−−−−−−−−−−√=9
C. (x+1)2+y2−−−−−−−−−−−√+(x−4)2+y2−−−−−−−−−−−√=81
D. (x−1)2+y2−−−−−−−−−−−√+(x+4)2+y2−−−−−−−−−−−√=81
4. Find the equation of a hyperbola that has foci at (−1,0) and (4,0), where the difference of the distances between each point on the ellipse and the two foci is 5.
A. (x+1)2+y2−−−−−−−−−−−√−(x−4)2+y2−−−−−−−−−−−√=25
B. (x−1)2+y2−−−−−−−−−−−√−(x+4)2+y2−−−−−−−−−−−√=5
C. (x−1)2+y2−−−−−−−−−−−√−(x+4)2+y2−−−−−−−−−−−√=25
D. (x+1)2+y2−−−−−−−−−−−√−(x−4)2+y2−−−−−−−−−−−√=5
Answer:
CABD
Step-by-step explanation:
HELPPPPPPPPPPPPPPP!!!!!!!
Answer:
I feel like the answers are 10 and -2
Step-by-step explanation:
Hope this helps! Have a great day and good luck! :)
A chocolate muffin recipe serves 12 people. An oatmeal raisin cookie recipe serves 36 people. A lemon cake recipe serves 16 people. Each recipe needs 2 eggs. The muffins need 3 cups of flour, while the cookies need 2 cups. The cake uses 2 lemons and the cookies use 1 cup of raisins and oatmeal. Each recipe needs 1 cup of sugar and milk. If we need to make cookies for 48 people, how much flour is needed?
To make cookies for 48 people, the recipe requires 6 cups of flour.
The oatmeal raisin cookie recipe serves 36 people and requires 2 eggs and 2 cups of flour. Since we need to make cookies for 48 people, we can calculate the amount of flour required as follows:
36 people → 2 cups of flour
1 person → (2 cups of flour) / (36 people) = (1/18) cups of flour
To make cookies for 48 people:
48 people × (1/18) cups of flour = 2.67 cups of flour
Therefore, to make cookies for 48 people, we need approximately 2.67 cups of flour.
Note: Since the result is a fraction of a cup, it is advisable to round up to the nearest whole number, so in this case, 3 cups of flour would be needed.
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suppose 40% of adults in the u.s. say they get their financial advice from family members. a random sample of 8 adults is selected. what is the probability at least 5 of the 8 say they get their financial advice from family members?
The probability that at least 5 of the 8 adults say they get their financial advice from family members can be calculated using binomial probability distribution.
Formula for binomial probability distribution is:P(X=k) = nCk * pk * (1-p)n-kwhere, P(X=k) is the probability of k successes in n independent trials, p is the probability of success in one trial, q=1-p is the probability of failure in one trial, nCk is the combination of k successes in n independent trials.
In this case, the probability of success is p=0.4 as 40% of adults in the US say they get their financial advice from family members.
Therefore, the probability of failure is q=1-0.4=0.6.So, P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)P(X=k) = nCk * pk * (1-p)n-kwhere, n=8, p=0.4, q=0.6For k = 5, nCk = 8C5P(X = 5) = 8C5 * (0.4)5 * (0.6)3= 0.2787
For k = 6, nCk = 8C6P(X = 6) = 8C6 * (0.4)6 * (0.6)2= 0.1960For k = 7, nCk = 8C7P(X = 7) = 8C7 * (0.4)7 * (0.6)1= 0.0575For k = 8, nCk = 8C8P(X = 8) = 8C8 * (0.4)8 * (0.6)0= 0.0030Therefore,P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2787 + 0.1960 + 0.0575 + 0.0030 = 0.5352
Hence, the probability that at least 5 of the 8 adults say they get their financial advice from family members is 0.5352.
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Given information that 40% of adults in the US say they get their financial advice from family members. We have to find the probability of at least 5 of the 8 say they get their financial advice from family members.
Hence, the required probability is 0.7530.
The probability of getting financial advice from family members is 40%. Let X be the number of people out of 8, who get their financial advice from family members. Here, X follows a binomial distribution with parameters n = 8 and p = 0.4. The probability of getting atleast 5 people getting their financial advice from family members is
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
Using binomial distribution formula, we get
P(X = x) = ${n\choose x}p^xq^{n-x}$
Where, n = 8, p = 0.4, q = 0.6
The probability of getting exactly 5 people out of 8 getting their financial advice from family members
P(X = 5) = ${8\choose 5} 0.4^5 (0.6)^{8-5}$
= 0.27869
The probability of getting exactly 6 people out of 8 getting their financial advice from family members
P(X = 6) = ${8\choose 6} 0.4^6 (0.6)^{8-6}$
= 0.29360
The probability of getting exactly 7 people out of 8 getting their financial advice from family members
P(X = 7) = ${8\choose 7} 0.4^7 (0.6)^{8-7}$
= 0.16493
The probability of getting exactly 8 people out of 8 getting their financial advice from family members
P(X = 8) = ${8\choose 8} 0.4^8 (0.6)^{8-8}$
= 0.01678
Therefore, the probability of getting at least 5 people getting their financial advice from family members is
P(X ≥ 5) = 0.27869 + 0.29360 + 0.16493 + 0.01678
= 0.7530
Hence, the required probability is 0.7530.
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With which information can you construct more than one triangle?
A the measurements of two angles
B the measurements of two angles and the length of the included side
C the measurements of all the angles
D the lengths of two sides and the measurement of the included angle
Answer:
B and D
Explanation:
1. Study Activity 1 on p.301. Then complete the following using the Sampling Distribution
of a Sample Proportion web app.
i. Simulate taking a random sample of 100 voters from a large population of voters of
whom 54% voted for Brown, and record the number out of 100 that voted for Brown.
ii. Report the proportion of your sample that voted for Brown
iii. Insert below the Data Distribution generated by the web app.
According to the question the following using the Sampling Distribution are as follows :
1. Study Activity 1 on p.301. Then complete the following using the Sampling Distribution of a Sample Proportion web app.
i. Simulate taking a random sample of 100 voters from a large population of voters of whom 54\% voted for Brown, and record the number out of 100 that voted for Brown.
ii. Report the proportion of your sample that voted for Brown
iii. Insert below the Data Distribution generated by the web app.
i. Simulate taking a random sample of 100 voters from a large population of voters of whom 54\% voted for Brown, and record the number out of 100 that voted for Brown.
ii. Report the proportion of your sample that voted for Brown: [Insert the proportion value here]
iii. Insert below the Data Distribution generated by the web app:
[Insert the data distribution plot here]
Note: The actual values for the proportion and data distribution will vary based on the simulation results obtained from the Sampling Distribution of a Sample Proportion web app.
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