It’s about The Normal distribution and other distributions Question: A study found that the mean amount of times cars spent in drive through a of a certain fast-food restaurant was 136.1 seconds. Assuming drive-through times are normally distributed with a standard deviation of 33 seconds, complete parts a through d below
A. What is the probability that a randomly selected car will get through the restaurant s drive thru in less than 82 seconds?
The probability that a randomly selected car will get through the restaurants drive thru in less than 82 seconds is?

439400

step-by-step explanation:

130^2 = 16900

16900*26 = 439400

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

it is 32.5

i hope i could u.