Anyone know the answer to this?????

HURRY ASAP!!!

9966 [12]

The answer you’re looking for is 5.00 :)

6 0

3 years ago

PLEASE HELP! 50 points and brainliest for correct answer and expatiation <br> What is m XY?

Delicious77 [7]

Answer:

99

Step-by-step explanation:

8 0

3 years ago

Pavlo invested some money in a bank account. The relationship between the elapsed time, ttt, in months, since Pavlo invested the

mart [117]

Answer: * =times\multiplication

<h2><u>every month the amount of money increases by a factor of 1.01</u></h2>

Explanation:

when t is 0 M(t) is 1000

when t is 1 M(t) is 1000 times 1.01

when t is 2 M(t) is 1000 times 1.01 squared which also looks like this:

1000* 1.01* 1.01

so in the end the factor is 1.01 because that is what is getting added each month (each month another 1.01 appears)

8 0

4 years ago

3.Find the value of each expression.

hoa [83]

Answer:

a) -3 b) -13 c) 60 d) 3/10 e) -112

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

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