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3 years ago

Find this area A) 492.5ft2 B) 508.2ft2 C) 515.3ft2 D)551.6ft2

Mathematics

1 answer:

kap26 [50]3 years ago

4 0

I think it’s A I’m sorry if I am wrong so I am pretty sure it’s A

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A group of 40 students from your school is part of the audience for a tv game show. The total number of people in the audience i

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18.75% chance they will be picked

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Write a quadratic equation with the following transformations.

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The answer I believe would be -6

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3 years ago

How to find the missing angle of this quadrilateral if the other angles are 108 degrees, 99 degrees, and 78 degrees?

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Alll angles in a quadrilateral add to 360

so
missing angle=x
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4 years ago

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

You want to know how many cups of birdseed are in each new bag. Which situation about Hayley’s birdseed best represents the equa

allochka39001 [22]

The answer is D. The +1 refers to the extra cup leftover from a previous bag. X represents how many cups are in each bag, and that is what you need to know. The three represents how many bags she bought, and the 25 represents how many cups of birdseed she had overall.

Hopefully this helps!

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3 years ago

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