The average score of all golfers for a particular course has a mean of 70and a standard deviation of 5.Suppose 100golfers played the course today. Find the probability that the average score of the 100golfers exceeded 71.Round to four decimal places.

0.9773

Step-by-step explanation:

Here, we start by calculating the z-scores statistic

Mathematically;

z-score = (x-mean)/SD/√n

From the question, we have;

x = 71, mean = 70, SD = 5 and n = 100

Plugging these values in the equation above, we have;

z-score = (71-70)/5/√100 = 1/5/10 = 1/0.5 = 2

So the probability we want to calculate is that;

P(z > 2)

This is obtainable from the standard normal distribution table

P(z > 2) = 0.97725 which is 0.9773 to 4 decimal places

72 fudge and 75 toffee

step-by-step explanation:

i hope this : )

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !