Changing the mean and standard deviation of a normal distribution by a moderate amount can greatly change the percent of observations in the tails. suppose a college is looking for applicants with sat math scores 780 and above.

(a) in a certain year, the scores of men on the math sat followed the n(531, 121) distribution. what percent of men scored 780 or better? (round your answer to two decimal places.) %

(b) women's sat math scores that year had the n(499, 112) distribution. what percent of women scored 780 or better? you see that the percent of men above 780 is almost three times the percent of women with such high scores. why this is true is controversial. (on the other hand, women score higher than men on the new sat writing test, though by a smaller amount. round your answer to two decimal places.)

560

hope this : )

-1   1/6.   you dont subtract the denominators

step-by-step explanation:

the degrees would remain the same ( the number to the left of the decimal point.

the multiply the decimal by 60 to get the minutes and use the numbers to the left of the decimal point.

then multiply the decimal portion of that by 60 for the seconds and round off as needed.

39 .956652

the 39 remains the degrees.

the minutes = 0.956652 x 60 = 57.39912

the minutes = 57

the seconds = 0.39912 x 60 = 23.95 = 24 ( rounded to nearest tenth)

so 39.956652 = 39 degrees, 57 minutes and 24 seconds

-75.159809

the -75 remains the degrees

the minutes = 0.159809 x 60 = 9.58854

the minutes = 9

the seconds = 0.58854 x 60 = 35.31, rounded to nearest tenth = 35.3 seconds

so -75.159809 = -75 degrees, 9 minutes and 35.3 seconds.

the first coordinate is north or south, since it it positive it is north.

the second number is east or west, because it is negative it is west.

39 degrees, 57 minutes and 24 seconds n

-75 degrees, 9 minutes and 35.3 seconds w