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mestny [16]
3 years ago
11

Plis help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
scoundrel [369]3 years ago
5 0

Answer:

Equation: 2x + 1 = 9x - 13

There are two chairs in each row.

Step-by-step explanation:

Looking at the problem, we see that Hue can form two rows of chairs with 1 left over, or 9 rows if she gets 13 more. With that information we can create the equation 2x + 1 = 9x - 13. Now let's solve for x.

2x + 1 = 9x - 13

<em>What we can do first is subtract 9x from both sides. This is done so all of the variables are on one side. Thus that becomes...</em>

2x + 1 - 9x = -13

<em>Now, combine -9x and 2x to get -7x.</em>

-7x + 1 = -13

<em>Now subtract 1 from both sides. This is done so we solve for x directly.</em>

-7x = -13 - 1

<em>Subtract -1 from -13 to get -14.</em>

-7x = -14

<em>Now divide both sides by -7.</em>

x = -14/-7

<em>Divide -14 by -7 to get 2. Remember, dividing two negatives gives you a positive!</em>

x = 2

With that, the equation is solved. There are 2 chairs in each row.

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Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.
zysi [14]

Answer:

a) 78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416

78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084

b) 78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989

78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d)   ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

Solving for n we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

And replacing the info we have:

n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55

Step-by-step explanation:

Part a

For this case we have the following data given

\bar X = 78.25 represent the sample mean for the customer order totals

\sigma =22.50 represent the population deviation

n= 40 represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be \alpha=0.1 and \alpha/2 = 0.05 and the critical value from the normal standard distirbution would be given by:

z_{\alpha/2}=1.64

And the confidence interval is given by:

\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}

And replacing we got:

78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416

78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989

78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

 ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

Solving for n we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

And replacing the info we have:

n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55

3 0
3 years ago
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