What is the coordinate (10,2) multiplied by two fifths equal

Simplify to a single power of 4:<br> 4^7/4^6

Temka [501]

Step-by-step explanation:

$\frac{ {4}^{7} }{ {4}^{6} }$

= 4

.......

4 0

3 years ago

Read 2 more answers

Which is the value of this expression when a=5 and k=-2

Zinaida [17]

Answer:

Option C is correct.

Step-by-step explanation:

We are given the expression:

$(\frac{3^2a^{-2}}{3a^{-1}})^k$

The value of a =5 and k = -2

Putting the values and solving

$=(\frac{3^2*5^{-2}}{3*5^{-1}})^-2\\=(\frac{3^{2-1}}{5^{-1+2}})^-2\\=(\frac{3^{1}}{5^{1}})^-2\\\\=(\frac{3}{5})^-2\\if \,\,a^{-1} \,\,then\,\, 1/a\\=\frac{(3)^{-2}}{(5)^{-2}}\\ Can\,\,be\,\,written\,\,as\\\\=\frac{(5)^{2}}{(3)^{2}} \\=\frac{25}{9}$

Option C is correct.

3 0

3 years ago

A researcher believes that the mean weight of competitive runners is about 140 pounds. A sample of 24 elite distance runners has

ExtremeBDS [4]

Answer:

$t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781$

$p_v =P(t_{(23)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ 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 mean is less than 140 pounds at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=136$ represent the sample mean

$s=11$ represent the sample standard deviation

$n=24$ sample size

$\mu_o =140$ represent the value that we want to test

$\alpha=0.01$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 140, the system of hypothesis would be:

Null hypothesis: $\mu \geq 140$

Alternative hypothesis: $\mu < 140$

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".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781$

P-value

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

$df=n-1=24-1=23$

Since is a one left tailed test the p value would be:

$p_v =P(t_{(23)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ 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 mean is less than 140 pounds at 5% of signficance.

4 0

2 years ago

Paulo's economics course requires two papers ? ?one long and one short ? ?throughout the semester. The number of pages, l l, in

mixer [17]

Answer:

27 pages.

Step-by-step explanation:

Let l be the number of pages in the long paper and s be the number of pages in the short paper.

We have been given that the total number of pages for both papers is 40. We can represent this information in an equation as:

$l+s=40...(1)$

We are also told that the number of pages in the long paper is one more than two times the number of pages in the short paper. We can represent this information in an equation as:

$l=2s+1...(2)$