Answer:
$28.50
Step-by-step explanation:
21 ÷ 14 = 1.5
19 · 1.5 = 28.50
Hope this helps!!
Answer:
7.64% probability that they spend less than $160 on back-to-college electronics
Step-by-step explanation:
Problems of normally distributed samples can be 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 237, \sigma = 54](https://tex.z-dn.net/?f=%5Cmu%20%3D%20237%2C%20%5Csigma%20%3D%2054)
Probability that they spend less than $160 on back-to-college electronics
This is the pvalue of Z when X = 160. So
![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{160 - 237}{54}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B160%20-%20237%7D%7B54%7D)
![Z = -1.43](https://tex.z-dn.net/?f=Z%20%3D%20-1.43)
has a pvalue of 0.0763
7.64% probability that they spend less than $160 on back-to-college electronics
The answer should be:
g <( or equal to)200 gal.
Answer:
![(x-7)^2+y^2=1](https://tex.z-dn.net/?f=%28x-7%29%5E2%2By%5E2%3D1)
Step-by-step explanation:
Hi there!
Equation of a circle:
where the circle is centered at (h,k) and the radius is r
<u>1) Plug in the given center (7,0)</u>
![(x-7)^2+(y-0)^2=r^2\\(x-7)^2+y^2=r^2](https://tex.z-dn.net/?f=%28x-7%29%5E2%2B%28y-0%29%5E2%3Dr%5E2%5C%5C%28x-7%29%5E2%2By%5E2%3Dr%5E2)
<u>2) Plug in the radius (1)</u>
![(x-7)^2+y^2=1^2\\(x-7)^2+y^2=1](https://tex.z-dn.net/?f=%28x-7%29%5E2%2By%5E2%3D1%5E2%5C%5C%28x-7%29%5E2%2By%5E2%3D1)
I hope this helps!