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Ludmilka [50]
2 years ago
7

Please help!! Thank you so much! (I hate school) :,)

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
5 0
The first and second one

(7x2)=14+5=19
(28/2)=14+5=19


Have a nice day!:)
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Is 7/8 - 2/5 about 1? Explain
ra1l [238]

Well we could start by making common denominators so we can subtract these two fractions.

7/8 = 35/40

2/5 = 16/40.

35/40 = 16/40 = 19/40.

19/40 = 0.475.

So about 1? Not really, since it rounds down to 0.

Close to 1? Not really, since it rounds down to 0.

Exactly? No.

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2 years ago
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Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the pop
Andreyy89

Answer:

The smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.

Step-by-step explanation:

The complete question is:

The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,103. A sample of n people will be selected at random from those living in the city. Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income. Round your answer up to the next largest whole number.

Solution:

The (1 - <em>α</em>)% confidence interval for population mean is:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

The margin of error for this interval is:

MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

The critical value of <em>z</em> for 90% confidence level is:

<em>z</em> = 1.645

Compute the required sample size as follows:

MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

      n=[\frac{z_{\alpha/2}\cdot\sigma}{MOE}]^{2}\\\\=[\frac{1.645\times 2103}{500}]^{2}\\\\=47.8707620769\\\\\approx 48

Thus, the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.

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IMPORTANT PLEASE HELP!!
Marianna [84]

Answer: I think it A

Step-by-step explanation:

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