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Sav [38]
1 year ago
6

Talia sells educational software for a yearly salary of $65,000 plus a 2.65% commission on her total sales. If Talia sells $2.5

million of software during one year, what are Talia’s earnings for that year?
Mathematics
1 answer:
Bad White [126]1 year ago
4 0

Answer:

$131250

Step-by-step explanation:

Talia's earnings during one year are equal to her base yearly salary ($65000) added to her commission. Now, to know the commission it is necessary to find the 2.65% of her total sales ($2.5 million)

Step 1. Divide the total sales or 2,500,000 by 100 considering this represents the total or 100%

2,500,000 ÷ 100 = 25000

Step 2. Multiply this number by the specific percentage (2.65)

25000 x 2.65 = 66250

Now add the commission and the yearly salary

65000 + 66250 = 131250

Hence, total earnings of Talia were $131250

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60+16+8 is 84

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2 years ago
It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the
JulijaS [17]

Answer:

1)  n=48  

2) n=298

3) n=426

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".  

p represent the real population proportion of interest

\hat p represent the estimated proportion for the sample

n is the sample size required (variable of interest)

z represent the critical value for the margin of error

The population proportion have the following distribution  

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

Part 1

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 90% of confidence, our significance level would be given by \alpha=1-0.90=0.10 and \alpha/2 =0.05. And the critical value would be given by:  

z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64  

The margin of error for the proportion interval is given by this formula:  

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

And on this case we have that ME =\pm 0.1 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:  

n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63  

And rounded up we have that n=48  

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7  

And rounded up we have that n=298  

Part 3

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 margin of error for the proportion interval is given by this formula:  

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

And on this case we have that ME =\pm 0.04 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:  

n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.96})^2}=425.22  

And rounded up we have that n=426  

3 0
2 years ago
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