5x/6 + 11/4 = 17/3 <br> Eliminate the fraction and solve for x

Show the reflection of the given points.Locate the points of the reflection in their proper position around the x-axis

amid [387]

Answer: I believe this is it

4 0

2 years ago

A private and a public university are located in the same city. For the private university, 1038 alumni were surveyed and 647 sa

Snezhnost [94]

Answer:

The difference in the sample proportions is not statistically significant at 0.05 significance level.

Step-by-step explanation:

Significance level is missing, it is α=0.05

Let p(public) be the proportion of alumni of the public university who attended at least one class reunion

p(private) be the proportion of alumni of the private university who attended at least one class reunion

Hypotheses are:

$H_{0}$ : p(public) = p(private)

$H_{a}$ : p(public) ≠ p(private)

The formula for the test statistic is given as:

z= $\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of public university students who attended at least one class reunion ( $\frac{808}{1311}=0.616$ )

p2 is the sample proportion of private university students who attended at least one class reunion ( $\frac{647}{1038}=0.623$ )

p is the pool proportion of p1 and p2 ( $\frac{808+647}{1311+1038}=0.619$ )

n1 is the sample size of the alumni from public university (1311)

n2 is the sample size of the students from private university (1038)

Then z= $\frac{0.616-0.623}{\sqrt{{0.619*0.381*(\frac{1}{1311} +\frac{1}{1038}) }}}$ =-0.207

Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.

6 0

3 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01

slavikrds [6]

Answer:

a) $z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

b) For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

$\hat p=\frac{174}{594}=0.293$ estimated proportion of yellow peas

$p_o=0.23$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

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 yellow peas is 0.23:

Null hypothesis: $p=0.23$

Alternative hypothesis: $p \neq 0.23$

When we conduct a proportion test we need to use the z statisitc, 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:

$z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

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

Since is a bilateral test the p value would be:

$p_v =2*P(z>3.649)=0.00026$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) Critical value

For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

5 0

3 years ago

If 350$ is placed in a savings account that earns 3.5% interest once a year, how much money will be in the account after 10 year

Anestetic [448]

472.50 because big maths

5 0

2 years ago

|2i+3j| is equal to:

IgorLugansk [536]

babe what are you doing on brainly?

8 0

2 years ago

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5x/6 + 11/4 = 17/3 Eliminate the fraction and solve for x