If the scores per round of golfers on the PGA tour are approximately normally distributed with mean 69.3 and standard deviation 2.58, what is the probability that a randomly chosen golfer's score is below 70 strokes

0.6069

Step-by-step explanation:

Given an approximately normal distribution:

Z = (x - mean) / standard deviation

Mean = 69.3 ; Standard deviation = 2.58

For :

P(x < 70)

Z = (70 - 69.3) / 2.58

Z = 0.7 / 2.58

Z = 0.2713

P(Z < 0.2713) = 0.6069

i leave the calculator work to you.

7. diameter 4 miles, circumference 4π miles

8. radius 0.2 miles, circumference 0.4π miles

[this is a bad problem because the actual radius of a car on the track is less than 0.2 miles so the answer is less than 0.4π miles]

9. radius 6 feet, circumference 12π feet

10. diameter 14 cm, circumference 14π cm

11. diameter 4.8 feet, circumference 4.8π feet

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.