Anyone want 2 trade brainliests with me or trade 50 points for a brainlist?

Plz help.will give brainliest

Oksi-84 [34.3K]

Answer:

x+2 x+2 x+2

Step-by-step explanation:

8 0

3 years ago

Solve each of the following equations for x showing the algebra you used to do it. Round each answer to the nearest ten-thousand

lyudmila [28]

11)46/7 thats the answer

4 0

4 years ago

Asia and taryn each had the same amount of money. after asia spent $14 and taryn spent $22, the ratio of asia's money to taryn's

blagie [28]

It is given in the question that

Asia and taryn each had the same amount of money. after asia spent $14 and taryn spent $22, the ratio of asia's money to taryn's money was 4 to 3.

Let , initially, they both have $x each .

So according to the given question ,

$\frac{x-14}{x-22} = \frac{4}{3}$

Cross multiplication

$3(x-14) = 4(x-22)
\\
3x-42 = 4x-88
\\
88-42 = 4x-3x
\\
x = 46$

Therefore , initially, they both have $46 each .

7 0

4 years ago

Can someone please helppp

ElenaW [278]

Answer:25%

Step-by-step explanation: you find how many pieces there are than you find how many shaded parts. You then put it in a fraction that time is by hundred. That should get you to you percentage. I hope this helped. I also this is the correct answer :)

6 0

3 years ago

Read 2 more answers

An e-commerce research company claims that 60% or more graduate students have bought merchandise on-line at their site. A consum

Elena-2011 [213]

Answer:

$z=\frac{0.55 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-0.913$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduate students show that only 44 students have ever done so is not significantly lower than 0.6

Step-by-step explanation:

Data given and notation

n=80 represent the random sample taken

X=44 represent the students that have bought merchandise on-line at their site

$\hat p=\frac{44}{80}=0.55$ estimated proportion of graduate students show that only 44 students have ever done so

$p_o=0.6$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of interest is lower than 0.6 or 60%, so then the system of hypothesis are.:

Null hypothesis: $p \geq 0.6$

Alternative hypothesis: $p < 0.6$

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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.55 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-0.913$

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 assumed for this case is $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduate students show that only 44 students have ever done so is not significantly lower than 0.6

5 0

3 years ago