The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds, What percent of players weigh between 180 and 220 pounds

57.62% of players weigh between 180 and 220 pounds

Step-by-step explanation:

When the distribution is normal, we use 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:

What percent of players weigh between 180 and 220 pounds

We have to find the pvalue of Z when X = 220 subtracted by the pvalue of Z when X = 180.

X = 220

has a pvalue of 0.7881

X = 180

has a pvalue of 0.2119

0.7881 - 0.2119 = 0.5762

57.62% of players weigh between 180 and 220 pounds

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x

answer:

c) the x-coordinate of the center of the circle increases by 3