The area is 864 square, or 6 square feet.
Answer:
Group B has the higher median by 0.5 inches. The IQR for Group A is smaller. IQR for Group B is 2 inches.
Answer:
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:

Find the probability that the diameter of a selected bearing is greater than 85 millimeters.
This is 1 subtracted by the pvalue of Z when X = 85. Then



has a pvalue of 0.7486.
1 - 0.7486 = 0.2514
0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.
1st) reflection over the point G
2nd) a dilation of 1/2 with fixed point G
3rd) a traslation 2 units left and 4 units up
Answer:
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IF Factor x4−10x2+25
x4−10x2+25
=(x2−5)(x2−5)
Answer:
(x2−5)(x2−5)
IF simplify
x4−10x2+25
There are no like terms.
Answer:
=x4−10x2+25