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k0ka [10]
3 years ago
15

A company tests all new brands of golf balls to ensure that they meet certain specifications. One test conducted is intended to

measure the average distance traveled when the ball is hit by a machine. Suppose the company wishes to estimate the mean distance for a new brand to within 1.9 yards with 90​% confidence. Assume that past tests have indicated that the standard deviation of the distances the machine hits golf balls is approximately 14 yards. How many golf balls should be hit by the machine to achieve the desired accuracy in estimating the​ mean?
Mathematics
1 answer:
docker41 [41]3 years ago
3 0

Answer:

385 golf balls

Step-by-step explanation:

margin of error = (z*)(s) / sqrt n

where z* = 1.96 with 5%/ 2 = 0.025 area in each tail

margin of error = (z*)(s) / sqrt n

1.2 yards = (1.96)(12 yards) / sqrt n

solve for n

n = 384.16

385 golf balls (always round up)

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Tyrone runs 5 mph how far can he travel in two hours and 30 minutes
zaharov [31]

Answer:

12.5 miles

Step-by-step explanation:

5x2.5=12.5 miles

5 0
3 years ago
Consider the following equation, bh+hr=25 when solving for r
Komok [63]
Solve for r.

You want to get r by itself on one side on the equal sign.

bh + hr = 25

Subtract bh from both sides.

hr = 25 - bh

Divide h on both sides.

r = 25 - bh / h

The two h's cancel each other out.

r = 25 - b

Hope this helps!
4 0
3 years ago
Read 2 more answers
The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day.
valina [46]

Answer:

A sample of 179 is needed.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.85}{2} = 0.075

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.075 = 0.925, so Z = 1.44.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

A previous study found that for an average family the variance is 1.69 gallon?

This means that \sigma = \sqrt{1.69} = 1.3

If they are using a 85% level of confidence, how large of a sample is required to estimate the mean usage of water?

A sample of n is needed, and n is found for M = 0.14. So

M = z\frac{\sigma}{\sqrt{n}}

0.14 = 1.44\frac{1.3}{\sqrt{n}}

0.14\sqrt{n} = 1.44*1.3

\sqrt{n} = \frac{1.44*1.3}{0.14}

(\sqrt{n})^2 = (\frac{1.44*1.3}{0.14})^2

n = 178.8

Rounding up

A sample of 179 is needed.

7 0
3 years ago
What is the interquartile range numbers 777,498,619,379,895,1256,1052​
Kay [80]

Answer:

554

Step-by-step explanation:

554  trust meh dude

8 0
2 years ago
Read 2 more answers
A 0.3 ml dose of a drug is injected into a patient steadily for 0.65 seconds. At the end of this time, the quantity, , of the dr
Darya [45]
Use the compound interest formula: A=P(1+i)^t.

P is the initial amount of the drug, 0.3ml.
i is - 0.0035.
t is in seconds.

You'll get:
A=0.3(1-0.0035)^t.

Sub in any value on t to find out how many ml are left t seconds after injection.

The 0.65 second injection time does not seem to be relevant as the question clearly states that the exponential decay starts AFTER the injection is completed.
4 0
3 years ago
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