Me. will mark brainliest. a. s.a. p
a club volleyball league allows only the top 5% of athletes who try out to be part of the team. if the team tryout scorecard has a mean of 250 and a standard deviation of 15, which of the following can be used as a minimum qualifying score to join the volleyball league?

a. 13
b. 237
c. 251
d. 274

D. 274

Step-by-step explanation:

A normal distribution of the scores is assumed. In the figure attached, the standard normal distribution table is shown.

If only top 5% of athletes are part of the team, then we need to find the value of the table which has a probability of 95%, that value is between 1.64 and 1.65, so we interpolate it as 1.645. The table was made for a variable with mean = 0 and standard deviation (sd) equal to 1, therefore to refer the result to our variable we compute:

1.645 = (x - mean)/sd

x = 1.645*sd + mean

x = 1.645*15 + 250 ≈ 274

So, 95% of the scores are below 274, then 274 is the minimum qualifying score

d is the correct answer

step-by-step explanation:

x= 8 because 3x = 3 * 8, and 3*8 = 24

"exterior angles = 360 / number of sides.

60 = 360 / sides

sides = 360 / 60

sides = 6"