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Dima020 [189]
2 years ago
11

PlEASE HELP ILL GIVE OUT BRAINLEIST

Mathematics
1 answer:
lesantik [10]2 years ago
5 0
Try 4 dollars and $.25
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1000add10000. 11000............
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Robin and Meg are traveling from New York to Florida, and plan to stop at three different campgrounds along the way. After choos
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You are correct. For each of the three possibilities for their second day, there are two possibilities for the third day.

The total number of possibilities is 3*2 = 6.

6 0
2 years ago
A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. A previous study indicates that the propo
never [62]

Answer:

A sample of 997 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

A previous study indicates that the proportion of left-handed golfers is 8%.

This means that \pi = 0.08

98% confidence level

So \alpha = 0.02, z is the value of Z that has a p-value of 1 - \frac{0.02}{2} = 0.99, so Z = 2.327.  

How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

This is n for which M = 0.02. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.02 = 2.327\sqrt{\frac{0.08*0.92}{n}}

0.02\sqrt{n} = 2.327\sqrt{0.08*0.92}

\sqrt{n} = \frac{2.327\sqrt{0.08*0.92}}{0.02}

(\sqrt{n})^2 = (\frac{2.327\sqrt{0.08*0.92}}{0.02})^2

n = 996.3

Rounding up:

A sample of 997 is needed.

3 0
3 years ago
What is 1/4 of 2000 1/5 of 1600 2/3 of 1965
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