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Lubov Fominskaja [6]
3 years ago
8

write a function that represents the situation: A population of 210,000 increases by 12.5% each year​

Mathematics
2 answers:
topjm [15]3 years ago
8 0
Y=2.5x+12,000 i think
Shkiper50 [21]3 years ago
3 0

Answer

y= 12.5x + 210,000

Step-by-step explanation:

This is a linear function because it is increasing constantly by 12.5 percent so it will me written as y=mx+b

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A cafeteria manager needs to know how many apples and bananas to to order. He asked students to choose either an apple or a bana
Usimov [2.4K]

Answer:

Step-by-step explanation:

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7 0
3 years ago
10.
Kay [80]

Answer:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.535 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.2 \leq \sigma \leq 2.8

And the best option would be:

A.  2.2 < σ < 2.8

Step-by-step explanation:

Information provided

\bar X=32.1 represent the sample mean

\mu population mean  

s=2.4 represent the sample standard deviation

n=83 represent the sample size  

Confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The degrees of freedom are given by:

df=n-1=83-1=82

The Confidence is given by 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, the critical values for this case are:

\chi^2_{\alpha/2}=62.132

\chi^2_{1- \alpha/2}=104.139

And replacing into the formula for the interval we got:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.535 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.2 \leq \sigma \leq 2.8

And the best option would be:

A.  2.2 < σ < 2.8

3 0
3 years ago
Can u guys help wit a little bit more questions what is 9.04-6.98 ​
Goryan [66]

Answer:

9.04-6.98=2.06

4 0
2 years ago
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