Answer:
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
The sketch is drawn at the end.
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.
Mean of 0°C and a standard deviation of 1.00°C.
This means that 
Find the probability that a randomly selected thermometer reads between −2.23 and −1.69
This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.
X = -1.69



has a p-value of 0.0455
X = -2.23



has a p-value of 0.0129
0.0455 - 0.0129 = 0.0326
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
Sketch:
What is the the x’s value?
Answer:
There will be <u>108</u> interactions for <u>9</u> children.
LCD is least common denominator. So looking at the denominators, their factors are...
First term x^2 - 5x + 4 = (x - )( x - ). We know this because the 4 (c term)is + so both signs are the same, and both are NEGATIVE because 5 (b term) is -
= (x - 4)(x - 1) because(-1)(-4) = 4 (c term) and -1 + -4 = -5 (b term)