Answer: 2.127
Step-by-step explanation:
For this problem, we have to divide the number 212.7 by 100 cookies.
212.7 ÷ 100 = 2.127
So, each cookie will require 2.127 ounces of dough.
Answer:
6:55 A. M. is the latest time he can leave his house
180
×.30
Is equals to 540
Remove the 0 and the answer is 54 miles.
Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
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 above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
![\mu = 502, \sigma = 115](https://tex.z-dn.net/?f=%5Cmu%20%3D%20502%2C%20%5Csigma%20%3D%20115)
The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:
X = 590:
![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{590 - 502}{115}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B590%20-%20502%7D%7B115%7D)
Z = 0.76
Z = 0.76 has a p-value of 0.7764.
X = 400:
![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{400 - 502}{115}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B400%20-%20502%7D%7B115%7D)
Z = -0.89
Z = -0.89 has a p-value of 0.1867.
0.7764 - 0.1867 = 0.5897 = 58.97%.
58.97% of students would be expected to score between 400 and 590.
More can be learned about the normal distribution at brainly.com/question/27643290
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