Please teach me how to work this out?!

I really need help guys idk how to do it

melomori [17]

Answer:

Jonah. Five million is a much higher number than 152,000. To put it mathematically -----> 5,000,000 > 152,000.

Step-by-step explanation:

7 0

3 years ago

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Eric is comparing the credit scores of his friends. The scores he gathered are found in the table below. 588 838 691 818 846 725

alukav5142 [94]

To solve this, we need to know how to find the mean of a set of data and how to find the median of a set of data.

To find the mean, or often called the average, we should add all of the values up, and then divide it by the number of values.

588+838+691+818+846+725+605+732+750 = 6593
6593/9=732.556

The problem tells us we should round to the nearest point, so our mean credit score is 733.

To find the median, we need to order the data from lowest to highest and find out which credit score(s) are right in the middle. If there are 2 in the middle, we simply should add them and divide by 2 to get our median. An easy way to do this is after you order them, you simply cross off one on each side until there is only 1 (or 2) left.

588 605 691 725 732 750 818 838 846
605 691 725 732 750 818 838
691 725 732 750 818
725 732 750
732

Since we only have one number in the middle, we are done with the median! We know our median is 732.

Now we simply need to compare them and subtract the lower one from the higher one.

Mean:733
Median: 732
733>732

We know the mean is bigger, so we should subtract the median from the mean.

733=732=1

Using the logic above, we can see that the mean is 1 point higher than the median.

8 0

3 years ago

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100 POINTS i need help fast!

mel-nik [20]

Answer:

Step-by-step explanation:

you multiple each hight and base

7 0

2 years ago

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In a study of 420,095 Danish cell phone users, 135 subjects developed cancer of the brain or nervous system (based on data from

Maksim231197 [3]

Answer:

$z=\frac{0.0003214 -0.00034}{\sqrt{\frac{0.00034(1-0.00034)}{420095}}}=-0.654$

$p_v =2*P(Z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.005$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that the true proportion not differs significantly from the specified value of 0.00034 or 0.034%.

Step-by-step explanation:

1) Data given and notation

n=420095 represent the random sample taken

X=135 represent the subjects developed cancer of the brain or nervous system (based on data from the Journal of the National Cancer Institute as reported in USA Today)

$\hat p=\frac{135}{420095}=0.0003214$ estimated proportion of subjects developed cancer of the brain or nervous system (based on data from the Journal of the National Cancer Institute as reported in USA Today)

$p_o=0.00034$ is the value that we want to test

$\alpha=0.005$ represent the significance level

Confidence=99.5% or 0.995

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the brain or nervous system at a rate that is different from the rate of 0.0340% :

Null hypothesis: $p=0.00034$

Alternative hypothesis: $p \neq 0.00034$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.0003214 -0.00034}{\sqrt{\frac{0.00034(1-0.00034)}{420095}}}=-0.654$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.005$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(Z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.005$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that the true proportion not differs significantly from the specified value of 0.00034 or 0.034%.

4 0

3 years ago

What is the rule for 1,3,5,7?

Anestetic [448]

You need to keep adding 2 to each number

7 0

3 years ago

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