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

the perimeter is increased by a factor of 2

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it