During a huge snowstorm in the White Mountains last year, it snowed 67.5 cm In one day. How much did it snow in meters?
2 answers:
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Answer:
A. length = 23 cm; width = 5 cm
Step-by-step explanation:
So if 4 times the width would be and 3 more than would be
, the equation would be
. If the perimeter is 56, then the new equation,
can be formed. This simplifies to
, then to
, then to
, and finally,
. So the width is 5, and now plug it into the first equation.
, becomes
, becomes
.
So the dimensions are: length = 23 cm, width = 5 cm.
*And just to be sure, which is correct.
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:
