Solve for r. l = prt

I really need help with this can you please answer the questions and say how you came up with it please I'm begging you

dangina [55]

A=30^2+ 2 x 30 x 26
A= 900 + 1560
A= 2460
so the surface area is 2460
v=1/3 x 30^2 x 26
v=1/3 x 900 x 26
v= 300 x 26
v=7800
so the volume is 7800
and the way the order of operations helped you is that you had to remember to always do the exponents before any of the other multiplication

4 0

4 years ago

Read 2 more answers

A long-distance athlete can run 1/2 kilometer in 3 minutes. How many kilometers can he run in an hour?

yarga [219]

He can run 10 km in an hour. 60 minutes divided by 3 is 20km. 20 would be the answer, but that's only if he runs 1km in 3 minutes. Since he runs .5km in 3 minutes, you need to divide the 20 by 2. Therefore, you have 10km in 1 hour

5 0

4 years ago

Find the area of the irregular shape

lys-0071 [83]

14 units^2 because you divide the shape into two parts find the area of each shape then add it together

7 0

3 years ago

Read 2 more answers

A painter took a digital photo of a 16-inch wide by 20-inch long painting to display on his website. The digital image is 2.5 in

xxTIMURxx [149]

The actual photo has an area of 16*20 inches, so 320 inches squared. The ratio between the digital and real photo is 16:20. On the other hand, the width to length of the digital photo is x:2.5. Cross multiply the two ratios and solve for x. 16/20*x/2.5 —> 20x = 40 —> x = 2. The width of the digital image is 2 inches, so multiplying the width times the length gives you a 5 inches squared area for the digital image. Answer: 5 in^2

7 0

4 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

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

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

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".

Solution to the problem

The population proportion have the following distribution

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

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

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

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

5 0

3 years ago