What is the outcome pls ifinrio

Is it possible for -52 over 7 to be simplified to 3 over 7

Rainbow [258]

As -52 and 7 have no common factors, -52/7 is a completely simplified improper fraction. If it were converted into a mixed number, it would be -8 4/7.

6 0

4 years ago

Franco is earns a weekly salary of

erik [133]

What is the rest of the question and you will probably have to multiply

8 0

4 years ago

A company claims that their new bottle holds 25 % more laundry soap. If their original container held 53 fluid ounces of soap, h

LiRa [457]

53 * .25 = 13.25, 53 + 13.25 = 66.25 fluid ounces

4 0

3 years ago

Read 2 more answers

What type of plate boundary joins mid ocean ridge segments quizzlet?

Anit [1.1K]

Mid ocean ridges are formed when two tectonic plates are drifting away from each other. This is known as a divergent or constructive plate boundary. Mid ocean ridges are formed as two oceanic plates move away from each other. The plates move as a result of convection cells in the mantle that generate convection currents. An example of a midocean ridge is the Mid Atlantic ridge where African plate drifts away from the north and south American plates.

4 0

4 years ago

Last year, a comprehensive report stated that 28% of businesses in the northeast of Ohio were considered highly profitable. This

Alik [6]

Answer:

$z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575$

$p_v =2*P(z>1.575)=0.115$

So the p value obtained was a very high value and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ 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 businesses were highly profitable is not significantly different from 0.28 or 28%.

Step-by-step explanation:

Data given and notation

n=50 represent the random sample taken

X=19 represent the businesses were highly profitable

$\hat p=\frac{19}{50}=0.38$ estimated proportion of businesses were highly profitable

$p_o=0.28$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 businesses were highly profitable is different from 0.28 or no, the system of hypothesis is.:

Null hypothesis: $p=0.28$

Alternative hypothesis: $p \neq 0.28$

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 required we can replace in formula (1) like this:

$z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575$

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 $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>1.575)=0.115$

So the p value obtained was a very high value and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ 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 businesses were highly profitable is not significantly different from 0.28 or 28%.

4 0

4 years ago