The answer is B
Have a good day!
7.8 billion/12 is how much per month:
7,800,000,000/12 = 650,000,000 candy per month.
To find out per person, divide the total amount of candies by the months in a year, then divide that amount to the population.
7,800,000,000/12 = 650,000,000
650,000,000/303,000,000 = 2.14 (Round down) = 2
Answers:
Per month: 650,000,000 candies
Per person: 2 candies
The decimal that’s equivalent to 28/100 is 0.28
Answer:
(0,-10)
Step-by-step explanation:
not to sure if this is what your asking for, sorry
Answer:
15.74% of women are between 65.5 inches and 68.5 inches.
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 = 64, \sigma = 1.5](https://tex.z-dn.net/?f=%5Cmu%20%3D%2064%2C%20%5Csigma%20%3D%201.5)
What percentage of women are between 65.5 inches and 68.5 inches?
This percentage is the pvalue of Z when X = 68.5 subtracted by the pvalue of Z when X = 65.5.
X = 68.5
![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{68.5 - 64}{1.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B68.5%20-%2064%7D%7B1.5%7D)
![Z = 3](https://tex.z-dn.net/?f=Z%20%3D%203)
has a pvalue of 0.9987
X = 65.5
![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{65.5 - 64}{1.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B65.5%20-%2064%7D%7B1.5%7D)
![Z = 1](https://tex.z-dn.net/?f=Z%20%3D%201)
has a pvalue of 0.8413
So 0.9987 - 0.8413 = 0.1574 = 15.74% of women are between 65.5 inches and 68.5 inches.