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0.9544 = 95.44% of scores lie between 220 and 380 points.
Normal distribution problems can be solved using the Z-score formula.
With a set of means and standard deviations, the Z-score for measure X is given by: After finding the Z-score, look at the Z-score table to find the p-value associated with that Z-score. This p-value is the probability that the value of the measure is less than X. H. Percentile of X. Subtract 1 from the p-value to get the probability that the value of the measure is greater than X.
We are given mean 300, standard deviation 40.
This means that µ= 300, σ = 40
What proportion of scores lie between 220 and 380 points?
This is the p-value of Z when X = 380 subtracted by the p-value of Z when X = 220.
X = 380
Z= (380-300)/40
Z= 2
Z=2 has a p-value of 0.9772.
X=300
Z= (220-380)/40
Z=-2
Z=-2 has a p-value of 0.9772.
0,9772 - 0,0228 = 0,9544
0.9544 = 95.44% of scores lie between 220 and 380 points.
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Answer:
20
Step-by-step explanation:
Given: F = 11, C = 11, E=?
By Euler's formula, we have:
Hence, the solid will have 20 edges.
The distance between the points A(-2, 3) and A'(2, 5) is 2√5 units if the point A(-2, 3) is translated using T: (x, y) → (x + 4, y + 2).
<h3>What is a distance formula?</h3>It is defined as the formula for finding the distance between two points. It has given the shortest path distance between two points.
The distance formula can be given as:
We have a point A(-2, 3)
After applying the translation:
T: (x,y) → (x + 4, y + 2)
A' = (-2+4, 3+2) = (2, 5)
The distance between A and A':
d = √20
or
d = 2√5 units
Thus, the distance between the points A(-2, 3) and A'(2, 5) is 2√5 units if the point A(-2, 3) is translated using T: (x, y) → (x + 4, y + 2).
Learn more about the distance formula here:
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