Plug in -5 for n
-5^3 + 3(-5)^2
-125 + 3(25) = -125 + 75 = -50
f(-5) = -50
Answer:
The area under the curve that represents the percent of women whose heights are at least 64 inches is 0.5.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X, or the area under the curve representing values that are lower than x. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X, which is the same as the area under the curve representing values that are higher than x.
In this problem, we have that:
![\mu = 64, \sigma = 2.75](https://tex.z-dn.net/?f=%5Cmu%20%3D%2064%2C%20%5Csigma%20%3D%202.75)
Find the area under the curve that represents the percent of women whose heights are at least 64 inches.
This is 1 subtracted by the pvalue of Z when X = 64.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{64 - 64}{2.75}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B64%20-%2064%7D%7B2.75%7D)
![Z = 0](https://tex.z-dn.net/?f=Z%20%3D%200)
has a pvalue of 0.5.
1 - 0.5 = 0.5
The area under the curve that represents the percent of women whose heights are at least 64 inches is 0.5.
Answer:
![x =18.0](https://tex.z-dn.net/?f=x%20%3D18.0)
Step-by-step explanation:
To solve this problem use the Law of cosine.
The law of cosine says that:
![c^2 = a^2 + b^2 -2abcos(C)](https://tex.z-dn.net/?f=c%5E2%20%3D%20a%5E2%20%2B%20b%5E2%20-2abcos%28C%29)
In this case we have that:
![c = x\\\\a=30\\\\b=16\\\\C=30\°](https://tex.z-dn.net/?f=c%20%3D%20x%5C%5C%5C%5Ca%3D30%5C%5C%5C%5Cb%3D16%5C%5C%5C%5CC%3D30%5C%C2%B0)
Therefore
![x^2 = 30^2 + 16^2 -2(30)(16)cos(30\°)](https://tex.z-dn.net/?f=x%5E2%20%3D%2030%5E2%20%2B%2016%5E2%20-2%2830%29%2816%29cos%2830%5C%C2%B0%29)
![x = \sqrt{30^2 + 16^2 -2(30)(16)cos(30\°)}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%7B30%5E2%20%2B%2016%5E2%20-2%2830%29%2816%29cos%2830%5C%C2%B0%29%7D)
![x = \sqrt{1156 -831.38}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%7B1156%20-831.38%7D)
![x = \sqrt{324.62}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%7B324.62%7D)
![x =18.0](https://tex.z-dn.net/?f=x%20%3D18.0)
Since she has 57 cans and 3 cans are allready devoted to the last row u must do 57-3 in or to get 54 cans total leftover. Then divide the number 54 by 9 since there are nine rows.so i think there will be six cans on each row of the shelf!
Answer with Step-by-step explanation:
Since we have given that
Mean = 150
Standard deviation = 8.8
An admissions officer at a large graduate school believes the mean is less than 150.
a)Determine the null and alternative hypotheses.
So, hypothesis :
![H_0:\mu =150\\\\H_a:\mu\neq 150](https://tex.z-dn.net/?f=H_0%3A%5Cmu%20%3D150%5C%5C%5C%5CH_a%3A%5Cmu%5Cneq%20150)
b)Explain what it would mean to make a Type I Error.
As we know that Type 1 error are those errors in which null hypothesis is supposed to be rejected but not get rejected.
So, here, mean is not less than 150 but in actual it is less than 150.
c)Explain what it would mean to make a Type II Error.
As we know that Type II error are those errors in which null hypothesis is not supposed to be rejected but it gets rejected.
So, mean is less than 150 but actually it is not less than 150.