Answer:
Step-by-step explanation:
Hello!
You need to construct a 95% CI for the population variance of the forces the safety helmets transmit to wearers.
The variable of interest is X: Force a helmet transmits its wearer when an external force is applied (pounds)
Assuming this variable has a normal distribution, the manufacturer expects it to have a mean of μ= 800 pounds and a standard deviation of σ= 40 pounds
A test sample of n=40 was taken and the resulting mean and variance are:
X[bar]= 825 pounds
S²= 2350 pounds²
To estimate the population variance per confidence interval you have to use the following statistic:
![X^2= \frac{(n-1)S^2}{Sigma^2} ~~X^2_{n-1}](https://tex.z-dn.net/?f=X%5E2%3D%20%5Cfrac%7B%28n-1%29S%5E2%7D%7BSigma%5E2%7D%20~~X%5E2_%7Bn-1%7D)
And the CI is calculated as:
[
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![X^2_{n-1;1-\alpha /2}= X^2_{39;0.975}= 58.1](https://tex.z-dn.net/?f=X%5E2_%7Bn-1%3B1-%5Calpha%20%2F2%7D%3D%20X%5E2_%7B39%3B0.975%7D%3D%2058.1)
![X^2_{n-1;\alpha /2}= X^2_{39;0.025}= 23.7](https://tex.z-dn.net/?f=X%5E2_%7Bn-1%3B%5Calpha%20%2F2%7D%3D%20X%5E2_%7B39%3B0.025%7D%3D%2023.7)
[
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]
[1577.45; 3867.09] pounds²
Using a confidence level of 95% you'd expect that the interval [1577.45; 3867.09] pounds² contains the value of the population variance of the force the safety helmets transmit to their wearers when an external force is applied.
I hope this helps!
Answer:
Yes it can, because −1.5 lies to the left of −0.5
On the surface of the pond
Answer:
amar mata
Step-by-step explanation:
:)
1) montly rate, r = 5% / 12 = 0.416% = 0.00416
2) capital, C = $ 450 = present value
3) time = 11 years => number of periods = 11 * 12 = 132
4) Formula: FV = C * (1 + r) ^ (number of periods)
5) Calculation: FV = $450 * ( 1 + 0.00416)^(132) = $ 779.07
Answer: $ 779.07
Answer:
Step-by-step explanation:
Given the functions z= (x+4y)e^y, x=u, and y =ln(v)
To get ∂z/∂u and ∂z/∂v, we will use the the composite rule formula:
∂z/∂u = ∂z/∂x•dx/du + ∂z/∂y•dy/du
∂z/∂x means we are to differentiate z with respect to x taking y as constant and this is gotten using product.
∂z/∂x = (x+4y)(0)+(1+4y)e^y
∂z/∂x = (1+4y)e^y
dx/du = 1
∂z/∂y = (x+4y)e^y+(x+4)e^y
dy/du = 0
∂z/∂u = (1+4y)e^y • 1 + 0
∂z/∂u = (1+4y)e^y
For ∂z/∂v:
∂z/∂v = ∂z/∂y• dy/dv
∂z/∂y = (x+4y)e^y+(x+4)e^y •(1/v)
∂z/∂y = {xe^y+4ye^y+xe^y+4e^y}•(1/v)
∂z/∂y = 2xe^y/v+4e^y(y+1)/v