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3 years ago

9

Par for a 9-hole golf course is 36. Jim was 3 under par for the first 5 holes. 4 over par for the next 2 holes, and 2 under par

for the last 2 holes. What was Jim's score?

1 answer:

Neporo4naja [7]3 years ago

4 0

Total the amounts he was over and under:

-3 + 4 -2 = -1

Now add that to the total par:

36 - 1 = 35

His score was a 35

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2 years ago

You panit 1/3 of a wall in 1/4 hour at that rate how long will it take you to paint one wall

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3 years ago

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MrRissso [65]

Answer:

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Step-by-step explanation:

3(h-7)+3=9

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3h-18=9

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3 years ago

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Use the standard norml distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justif

dalvyx [7]

Answer:

E) we will use t- distribution because is un-known,n<30

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Step-by-step explanation:

<u>Step:-1</u>

Given sample size is n = 23<30 mortgage institutions

The mean interest rate 'x' = 0.0365

The standard deviation 'S' = 0.0046

the degree of freedom = n-1 = 23-1=22

99% of confidence intervals $t_{0.01} =2.82$ (from tabulated value).

$The mean value = 0.0365$

$x±t_{0.01} \frac{S}{\sqrt{n-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{23-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{22} }$

$0.0365±2.82 \frac{0.0046}{4.690 }$

using calculator

$0.0365±0.00276$

Confidence interval is

$(0.0365-0.00276,0.0365+0.00276)$

$(0.0338,0.0392)$

the mean value is lies between in this confidence interval

(0.0338,0.0392).

<u>Answer:-</u>

<u>using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.</u>

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