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densk [106]
3 years ago
11

Can y’all help me with 7 - 12 showing work. Will be giving Brainliest:)) and for 13 and 14 I need help too :)) Help please :((((

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
4 0
7= 16 lb >16 Oz 16x16=256
8= 1500 lb < 2T 2000+2000
9= 3T>5999 lb 2000x3
10= 1600 Oz > 10 lb 1600÷16=100
11= 19 lb>300 Oz 19x16=304oz
12= 8 oz= 1/2 lb 16 ounces = 1 pound 1/2=8
13= pounds=6000 tons= 1, 1.5
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Suppose we want to choose 5 letters, without replacement, from 9 distinct letters. (If necessary, consult a list of formulas.) (
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This was someone else's work not mine sorry here's creditssss :))

brainly.com/question/15145413

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There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f
Ann [662]

Answer:

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0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

The estimated population proportion for this case is:

\hat p = \frac{350}{500}=0.7

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647

0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

7 0
2 years ago
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