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2 years ago

Jane needs $20 to buy her radio. She has saved $15. What percent of the cost of the radio has she saved?

Mathematics

2 answers:

aivan3 [116]2 years ago

8 0

Answer: 75%

Step-by-step explanation:

15/20 = 3/4 or 75%

zepelin [54]2 years ago

6 0

Answer:

I think she saved 5 percent.

Step-by-step explanation:

I am not 100% sure tho

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A company's current stock price is $50.00 and its most recent dividend was $2.00 per share. Since analysts estimate the company

4vir4ik [10]

Answer:

Expected rate of return =7.1% (Approx.)

Step-by-step explanation:

Given:

Current stock price = $50

Divided d = $2

Growth rate g = 5 %

Find:

Expected rate of return

Computation:

Expected rate of return = D(1+g)/Current Price + g

Expected rate of return = [2(1+5%)/50] + 5%

Expected rate of return =7.1% (Approx.)

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yaroslaw [1]

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3 years ago

You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample

Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=160)$

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{160}{\sqrt{n}})$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 1.96$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{1.96(160)}{78.4})^2 =16$

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And now we can replace the new value of Me and see what we got, like this:

$n=(\frac{1.96*160}{60})^2 =27.32$