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pishuonlain [190]
3 years ago
5

An abolitionist supports which of the following causes?

Mathematics
2 answers:
Aliun [14]3 years ago
8 0
Ending slavery and ending women sufferage
skelet666 [1.2K]3 years ago
8 0
Ending woman suffering and ending slavery
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Cereal is on sale this week. Is it a better buy to get the 10-ounce box for $1.76 or the 14-ounce box for $2.42?
Veseljchak [2.6K]

Answer:

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Step-by-step explanation:

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2 years ago
5 + 5 + ?<br> Im Need To Know Please ; ]
-BARSIC- [3]

Answer:

5 plus 5 would equate to the variable of 10 :)

Step-by-step explanation:

5+5=10

1+1+1+1+1 + 1+1+1+1+1 = 10

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3 years ago
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Citrus2011 [14]
The answer the the question is A.
5 0
3 years ago
How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide
kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

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

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

The margin of error of a (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

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

The information provided is:

<em>σ</em> = $60

<em>MOE</em> = $2

The critical value of <em>z</em> for 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the sample size as follows:

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

       n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}

          =[\frac{1.96\times 60}{2}]^{2}

          =3457.44\\\approx 3458

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

8 0
2 years ago
How do I ask a question about a graph
Tatiana [17]

Answer:

take a screen shot and then post it

Step-by-step explanation:

mark me brainlist

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2 years ago
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