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4 years ago

PLEASE HELP!!!!

Mathematics

1 answer:

LekaFEV [45]4 years ago

4 0

Answer:

500

Step-by-step explanation:

It's a box with a square base, so let's say the width and length are x and the height is y.

The surface area of the box without the top is:

A = x² + 4xy

300 = x² + 4xy

The volume of the box is:

V = x²y

Solve for y in the first equation and substitute into the second:

y = (300 − x²) / 4x

V = x² (300 − x²) / 4x

V = x (300 − x²) / 4

V = 75x − ¼ x³

To optimize V, find dV/dx and set to 0:

dV/dx = 75 − ¾ x²

0 = 75 − ¾ x²

x = 10

So the volume of the box is:

V = 75x − ¼ x³

V = 500

The maximum volume is 500 cm³.

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Maria studied the traffic trends in India. She found that the number of cars on the road increases by 10% each year. If there we

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Step-by-step explanation:

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4 years ago

The Pew Research Center has conducted extensive research on the young adult population (Pew Research website, November 6, 2012).

Rudik [331]

Answer:

a) $0.93 - 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.908$

$0.93 + 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.952$

The 95% confidence interval would be given by (0.908;0.0.952)

b) $0.21 - 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.163$

$0.21 + 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.257$

The 99% confidence interval would be given by (0.163;0.0.257)

c) The margin of error for part a is:

$ME= 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.0224$

And for part b is:

$ME=2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.0470$

So then the margin of error is larger for part b.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.93 - 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.908$

$0.93 + 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.952$

The 95% confidence interval would be given by (0.908;0.0.952)

Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.21 - 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.163$

$0.21 + 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.257$

The 99% confidence interval would be given by (0.163;0.0.257)

Part c

The margin of error for part a is:

$ME= 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.0224$

And for part b is:

$ME=2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.0470$

So then the margin of error is larger for part b.

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shtirl [24]

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2 years ago