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:
1.5 hours=60+30=90 minutes
30 pages : 90 minutes = 0.3 pages/minute
The numbers are 4 and 2! 4 plus 2 equals 6! and 6 plus 4 is equal 10!
Answer:
x= 1 :))
Step-by-step explanation:
Using the distributive property we know we distribute by multiplying the outside values to those inside of the parentheses
outside values: 2
inside values: 5 and r
now multiply out inside values by the outside values
2x5 = 10
2xr = 2r
now plug these values back in for the inside values and take away the outside value and the parentheses
so 2(5+r) using the distribution property is 10+2r