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3 years ago

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What is the cost of 4 pairs of Nikes if each pair is $365.68 along with 120 peaches that each cost $9.00. After he lost 3 pairs

of his shoes and he ate 89 peaches? please help!! giving away 20 point(s)

2 answers:

Monica [59]3 years ago

5 0

Answer:

this was tricky but I think I have the answer. 1046.68 dollar s

Andreas93 [3]3 years ago

3 0

One cost of a pair of Nike's is $365.68 so to find the cost of 4 pairs, we multiply 365.68 by 4 which gets us $1462.72

And he is fat

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How much does the atmosphere weigh over each square meter of surface? express this in tonnes (units of 1000 kg). recall the conv

Anettt [7]

Atmospheric pressure = 101,325 Newtons / m²
Weight per meter square = 101,325 Newtons
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Mass per meter square = 10.3 tonnes per m²

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3 years ago

If logx √2 = 1/6<br> , then x =?

Kaylis [27]

Step-by-step explanation:

logx√2=1/6

<=> logx(2^1/2)=1/6

<=>1/2.logx(2)=1/6

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Sorry I forgot the formula in the last step

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3 years ago

Read 2 more answers

6.Suppose the Gallup Organization wants to estimate the population proportion of those who think there should be a law that woul

drek231 [11]

Answer:

A sample of $n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$ is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

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

In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.

This means that $n = 1012, \pi = \frac{374}{1012} = 0.3696$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.

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

$E = 1.96\sqrt{\frac{0.3696*0.6304}{n}}$

$E\sqrt{n} = 1.96\sqrt{0.3696*0.6304}$

$\sqrt{n} = \frac{1.96\sqrt{0.3696*0.6304}}{E}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$

$n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$

A sample of $n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$ is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

4 0

3 years ago

Answer all parts of number 1 thank you.

Scorpion4ik [409]

Hi!

1.)

a:A u B=

$(1.0. - 5. \frac{1}{2} .e)$

(The same elements are written once.)

b:A u C

$(1.0 - 5. \frac{1}{2} .2)$

c:B u C

$( - 5 .1.e.0.2)$

Note:I used dots to mean commas .

Good work!

7 0

1 year ago

The mean of a certain set of measurements is 27 with a standard deviation of 14. The distribution of the

Talja [164]

Answer:

Exactly 16%.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The mean of a certain set of measurements is 27 with a standard deviation of 14.

This means that $\mu = 27, \sigma = 14$

The proportion of measurements that is less than 13 is

This is the p-value of Z when X = 13, so:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{13 - 27}{14}$

$Z = -1$

$Z = -1$ has a p-value of 0.16, and thus, the probability is: Exactly 16%.

8 0

2 years ago