Help please, 100 points

What is 6 divided by 3647? I need to show work. Much appreciated.

Andre45 [30]

the answer is 607 R 5

3 0

3 years ago

The same number of people live on the islands Beautiful Sunrise and Gorgeous

aev [14]

Using the percentage concept, it is found that 75% of the population of Gorgeous Sunset is on Beautiful Sunrise now.

<h3>What is a percentage?</h3>

The percentage of an amount a over a total amount b is given by a multiplied by 100% and divided by b, that is:

$P = \frac{a}{b} \times 100\%$

In this problem, we have that:

We consider that the population of both Beautiful Sunrise and Gorgeous Sunset islands is of x.

There is a fiesta at Beautiful Sunrise, and a number a of people from Gorgeous Sunset are coming, hence, there will be x + a people at Beautiful Sunrise and x - a people t Gorgeous Sunset.

The percentage of people from Gorgeous Sunset is on Beautiful Sunrise now is:

$P = \frac{a}{x} \times 100\%$

Now the number of people on Beautiful Sunrise is seven times the number of people on Gorgeous Sunset, hence:

$\frac{x + a}{x - a} = 7$

We can find a <u>as a function of x</u> to find the percentage:

$\frac{x + a}{x - a} = 7$

$7(x - a) = x + a$

$7x - 7a = x + a$

$8a = 6x$

$a = \frac{3x}{4}$

Then, the percentage is:

$P = \frac{a}{x} \times 100\%$

$P = \frac{\frac{3x}{4}}{x} \times 100\%$

$P = \frac{3}{4} \times 100\%$

$P = 75\%$

75% of the population of Gorgeous Sunset is on Beautiful Sunrise now.

You can learn more about the percentage concept at brainly.com/question/10491646

7 0

2 years ago

A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa

Norma-Jean [14]

Answer:

a) Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

b) $df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.

c) Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation

$\bar X=18.3$ represent the sample mean

$s=3.72$ represent the sample standard deviation

$n=25$ sample size

$\mu_o =15$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b: P-value and conclusion

The first step is calculate the degrees of freedom, on this case:

$df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.

Part c

Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0

3 years ago

LaSanda bought 20 shares of stock for $175. She sold the stock for a total profit of $25. What was the selling price of each sha

podryga [215]

So if she bought 20 share of stock for $175 that means that she paid $8.75 for each stock so if she made $25 you add then you divide

175÷20 = $8.75
175+$25=$200 she sold them for
$200 ÷ 20= $10

4 0

3 years ago

What is the mean of this data set: 18, 15, 33, 34, 55

WARRIOR [948]

Answer:

18

Step-by-step explanation:

33-15=18 I belive that is the mean

8 0

3 years ago