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3 years ago

PLEASE HELP ASAP! BRAINLIEST TO BEST/RIGHT ANSWER!

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

7 0

In this problem, we are dealing with simplifying expressions with exponents in the denominator. When we have exponents in the denominator, one thing that we want to be sure of is that there are not any negative exponents and that the numerators and denominators are completely simplified.

Our expression is 5/x^-2y^5

The first thing we notice is that we have a negative exponent in the denominator. This is easily solved if we switch the term with the negative exponent into the numerator.

Our expression now looks like this: 5x^2/y^5

Now, there aren't any other terms that cancel out, so our simplified expression is 5x^2/y^5.

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15 is 60% of what number?

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15 is 60 percent of 25. 15x100/60 and it should be 25.

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A survey was conducted in the United Kingdom, where respondents were asked if they had a university degree. One question asked,

katovenus [111]

Answer:

$z=\frac{0.121-0.0892}{\sqrt{0.101(1-0.101)(\frac{1}{373}+\frac{1}{639})}}=1.620$

$p_v =2*P(Z>1.620)=0.105$

If we compare the p value and using any significance level for example $\alpha=0.05$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the two proportions are not statistically different at 5% of significance

Step-by-step explanation:

Data given and notation

$X_{1}=45$ represent the number of correct answers for university degree holders

$X_{2}=57$ represent the number of correct answers for university non-degree holders

$n_{1}=373$ sample 1 selected

$n_{2}=639$ sample 2 selected

$p_{1}=\frac{45}{373}=0.121$ represent the proportion of correct answers for university degree holders

$p_{2}=\frac{57}{639}=0.0892$ represent the proportion of correct answers for university non-degree holders

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportions are different between the two groups, the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{45+57}{373+639}=0.101$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.121-0.0892}{\sqrt{0.101(1-0.101)(\frac{1}{373}+\frac{1}{639})}}=1.620$

Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test.

Since is a one side test the p value would be:

$p_v =2*P(Z>1.620)=0.105$

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Answer:

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Answer:

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Step-by-step explanation:

I put the answer in decimal form and rounded up from .0706, but pi form in included in the pic. I hope this helped! Please leave Brainliest if it did and is right.

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