Answer the math question please

It takes 41 minutes for 9 people to paint 9 walls. how long does it take 14 men to paint 14 walls

denis23 [38]

41, because the ratio is the same

5 0

3 years ago

Nathan goes to the store and buys 2 shorts and 4 shirts for 70$. He later goes to the same store and buys 3 shorts and 3 shirts

stiks02 [169]

Answer:

Each item of clothing costs $12.50

Please Mark brainliest

Step-by-step explanation:

8 0

3 years ago

the cost to operate a certain aircraft is $255 per hour the algebraic expression 255t gives the total cost to operate the aircra

Temka [501]

Answer:

Total cost for 2.80 hours = ($255/hr)(2.80 hrs) = $714.

Step-by-step explanation:

Multiply this hourly rate by the elapsed time (2.80 hours):

Total cost for 2.80 hours = ($255/hr)(2.80 hrs) = $714.

3 0

3 years ago

A spotlight has a beam that travels 93 feet and covers an area intercepted by an 76 degrees angle, as shown below

Novay_Z [31]

Answer:7330 feet squared

Step-by-step explanation:

I think its right

4 0

3 years ago

Suppose a poll is taken that shows that 765 out of 1500 randomly selected, independent people believe the rich should pay more

Zanzabum

Answer:

$z=\frac{0.51 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1500}}}=0.775$

$p_v =P(z>0.775)=0.219$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ 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 interest is not significantly higher than 0.5

Step-by-step explanation:

Data given and notation

n=1500 represent the random sample taken

X=765 represent the successes

$\hat p=\frac{765}{1500}=0.51$ estimated proportion of successes

$p_o=0.5$ 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 is higher than 0.5:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

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

$z=\frac{0.51 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1500}}}=0.775$

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 right tailed test the p value would be:

$p_v =P(z>0.775)=0.219$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ 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 interest is not significantly higher than 0.5

8 0

3 years ago