Raj bought 300 mangoes at 10 for Rs 100 and sold them at o profit of 25%. At what rate did he sell per 15 mangoes.

From the trail head, the waterfall is 4/5 mile, and then the eagle’s nest is 3/4 mile farther. From there, it is 1 3/8 miles bac

Anvisha [2.4K]

Answer:

Need more info

Step-by-step explanation:

5 0

3 years ago

Please help! Giving brainliest

lianna [129]

Answer:

5y^2-12y+21-73/2y+3

6 0

2 years ago

4. Tristan has a total of 30 dimes & nickels. The total value

adelina 88 [10]

Answer:

x/2 + x = 30 -> $2.50

20/2 + 20 = 30

X = 20, therefore:

dimes = 20

nickels = 10

4 0

2 years ago

Read 2 more answers

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

Please answer please

Mariulka [41]

Answer:

that's wrong

Step-by-step explanation:

so yesThese icons are frequently found in the Quick Launch area. (Choose all that apply) Show Desktop icon My Documents directory Recycle Bin Windows Media Player Internet Explorer. best new y jk lol idk what that is XD

6 0

3 years ago