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:
Answer:
ig 50 meters long by 25 meters wide
Step-by-step explanation:
if wrong pls forgive me
mark me brainliest pls
Answer:
1 | 38
2 | 76
3 | 114
5 | 190
1/2 | 19
2/3 | 25.33
Step-by-step explanation:
I am assuming that you need to multiply the area by the scale factors
Y=2(4)+4
=12 gallons in the tank so far
50-12
=38 gallons to fill
Answer: 400=12.50x+50
400-50=12.50x
350=12.50x
350÷12.50=x
28=x
Step-by-step explanation: