You paint 1/3 of a wall in 1/4 of an hour at this rate how long will it take for you to paint one whole wall

The average median income for a dentist is $142,740. The average median income for a medical and health service manager is $86,4

KiRa [710]

A: For the Dentist, their average income in 30 years is $4,282,200

B: For the Manager, their average income in 30 years is $2,592,000

C: The difference in both their earnings is $1,690,200

Work: So, I just multiplied their average earnings by 30, and that's what their average income will be in 30 years. For the difference, I subtracted the higher earning by the lower earning, and that's the difference.

I hope this helps you. :)

7 0

3 years ago

ANSWER FAST AND I WILL PUT BRAINLIEST... Minnie graphs the temperature over a 24-hour period. Which type of graph should she use

sergij07 [2.7K]

As the temperature may remain constant over some period of time or may change at every moment So a continuous graph will be the best way to show the temperature over 24 hour period.

So the answer should be

A continuous graph

3 0

3 years ago

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I need help plss :)!

OLEGan [10]

3wx^2

The third choice is the answer.

Answer :C

7 0

3 years ago

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The price of an item has been reduced by 55%. The original price was $85.

Shkiper50 [21]

Answer:

38.25

Step-by-step explanation:

Multiply 85 by 55 and you get 4675 then divide by 100.

Then you will get 46.75. Then subtract 46.75 from 85

4 0

3 years ago

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Use the Statistical Applet: P ‑Value for a Test of One Proportion to answer the question.You have taken a sample of ????=1000 in

umka21 [38]

Answer:

$z=\frac{0.785 -0.7}{\sqrt{\frac{0.7(1-0.7)}{750}}}=5.08$

$p_v =P(z>5.08)=1.88x10^{-7}$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is highr than 0.67 or 67% .

Step-by-step explanation:

1) Data given and notation

n=20 represent the random sample taken

X=15 represent the people who eat breakfast

$\hat p=\frac{15}{20}=0.75$ estimated proportion of people who eat breakfast

$p_o=0.67$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 67%:

Null hypothesis: $p\leq 0.67$

Alternative hypothesis: $p > 0.67$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.785 -0.7}{\sqrt{\frac{0.7(1-0.7)}{750}}}=5.08$

4) 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 next step would be calculate the p value for this test.

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

$p_v =P(z>5.08)=1.88x10^{-7}$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is highr than 0.67 or 67% .

6 0

3 years ago