A golfer hits a ball from a starting elevation of 4 feet with a velocity of 70 feet per second down to a green with an elevation
1 answer:
Answer:
t = 4.5 seconds
Step-by-step explanation:
It is given that, a golfer hits a ball from a starting elevation of 4 feet with a velocity of 70 feet per second down to a green with an elevation of −5 feet.
The number of seconds t it takes the ball to hit the green can be represented by the equation as follows :
......(1)
We need to find time taken by the ball to land on the green. Solving the quadratic equation (1) we get :
t = -0.125 s and t = 4.5 s
Ignoring t = -0.125 s, we get the time taken by the ball to land on the green is 4.5 seconds.
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Using the z-distribution, it is found that the 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
We are given the <u>standard deviation for the population</u>, hence, the z-distribution is used. The parameters for the interval is:
- Sample mean of
- Population standard deviation of
- Sample size of
.
The margin of error is:
In which z is the critical value.
We have to find the critical value, which is z with a p-value of , in which
is the confidence level.
In this problem, , thus, z with a p-value of
, which means that it is z = 1.96.
Then:
The confidence interval is the <u>sample mean plus/minutes the margin of error</u>, hence:
The 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
A similar problem is given at brainly.com/question/22596713