The graph below shows a probability distribution. The answer choices are below. Thanks! :)

What is 5(4x+3)= simplified

cluponka [151]

5(4x+3)=0

20x+15=0

20x = -15

$x = \frac{ - 15}{20}$

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

4 0

3 years ago

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A rectangular pool is 6 meters wide and 8 meters long. If you swim diagonally across the pool, how many meters would you be swim

Vitek1552 [10]

If you swim diagonally across the rectangular pool, the distance you swim is 10 meters.

Given the following data:

Width of rectangle = 6 meters

Length of rectangle = 8 meters

To determine the distance you swim in meters, we would apply Pythagorean's theorem since the width is along the x-axis while the length is along the y-axis.

Note: The diagonal side of the rectangular pool represents the hypotenuse.

Mathematically, Pythagorean's theorem is given by the formula:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

Substituting the given parameters into the formula, we have;

$Hypotenuse^2 = 6^2 + 8^2\\\\Hypotenuse^2 = 36+64\\\\Hypotenuse^2 = 100\\\\Hypotenuse = \sqrt{100}$

Hypotenuse = 10 meters.

Read more here: brainly.com/question/18890335

4 0

2 years ago

Complete the synthetic division problem below.

miskamm [114]

Step-by-step explanation:

c

6 0

3 years ago

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Can somebody plz answer this math problem correctly thanks! WILL MARK BRAINLIEST :D

Soloha48 [4]

Answer:

Ellen and Blake answered 20 questions correctly since Ellen had 0.8 correct and that is 80% same with Blake

Stephen scored a .84 or an 84%

I hope this helps :)

Step-by-step explanation:

plz mark B R A I N L I E S T

3 0

2 years ago

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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