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Ganezh [65]
3 years ago
8

Each of 16 students will give a 1 1/5 minute speech in English class. How long will it take to give the speeches?

Mathematics
1 answer:
Gwar [14]3 years ago
7 0

Answer:

35 and 1/5

Step-by-step explanation:

you multiply.

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Yes this is the best kind of question
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Tristan made 98 bracelets in 7 hours and Denita made 56 bracelets in 4 hours. These rates are equivalent because they have the s
dexar [7]

Answer:

A) unit rate

Hope this helps

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3 years ago
5. what is the theoretical probability that a coin toss results in one head and one tail showing?
Alenkasestr [34]
There are two sides (heads and tails). So one head would be 1/2, and one tail would also be 1/2.
4 0
3 years ago
what is the volume of the cone with radius 6 inches and height 10 inches express the answer in terms of pi
Vanyuwa [196]

Answer:

V = (1/3)(π)(6 in)²(10 in) = 120π in³

Step-by-step explanation:

The volume of a cone of radius r and height h is V = (1/3)πr²h.

Here, that volume is V = (1/3)(π)(6 in)²(10 in) = 120π in³

6 0
3 years ago
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Quit smoking: In a survey of 444 HIV-positive smokers, 202 reported that they had used a nicotine patch to try to quit smoking.
Flauer [41]

Answer:

The p-value of the test is of 0.0287 < 0.05(standard significance level), which means that it can be concluded that less than half of HIV-positive smokers have used a nicotine patch.

Step-by-step explanation:

Test if less than half of HIV-positive smokers have used a nicotine patch:

At the null hypothesis, we test if the proportion is of at least half, that is:

H_0: p \geq 0.5

At the alternative hypothesis, we test if the proportion is below 0.5, that is:

H_1: p < 0.5

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.5 is tested at the null hypothesis:

This means that \mu = 0.5, \sigma = \sqrt{0.5*(1-0.5)} = 0.5

In a survey of 444 HIV-positive smokers, 202 reported that they had used a nicotine patch to try to quit smoking.

This means that n = 444, X = \frac{202}{444} = 0.455

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.455 - 0.5}{\frac{0.5}{\sqrt{444}}}

z = -1.9

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.455, which is the p-value of z = -1.9.

Looking at the z-table, z = -1.9 has a p-value of 0.0287.

The p-value is of 0.0287 < 0.05(standard significance level), which means that it can be concluded that less than half of HIV-positive smokers have used a nicotine patch.

5 0
3 years ago
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