Answer:
The child's reading level is at the 3.4th percentile.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

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 p-value, we get the probability that the value of the measure is greater than X.
You do some research and determine that the reading rates for their grade level are normally distributed with a mean of 90 words per minute and a standard deviation of 24 words per minute.
This means that 
You find an individual that reads 46.4 word per minute. At what percentile is the child's reading level?
The percentile is the p-value of Z when X = 46.4, multiplied by 100. So



has a p-value of 0.034.
0.034*100 = 3.4
The child's reading level is at the 3.4th percentile.
SOLUTION:
Step 1:
In this question, we are given the following:
A cup of coffee contains 100 mg of caffeine, which leaves the body at a continuous rate of 17% per hour.
Part A) Write a formula for the amount, A mg, of caffeine in the body t hours after drinking a cup of coffee.
Step 2:
The details of the solution are as follows:
The graph of:

is as follows:
The amount of the markup is 20% of 25.15
EQUATION 1: 20/100 * 25.15
ANSWER TO EQUATION 1: 503/100
Then we have divide 503 and 100
EQUATION 2: 503/100
FINAL ANSWER: $5.03
The amount of the markup is $5.03
Replace x in the equation and see which one equals the y values:
The answer is f(x) = 2x - 1