What’s 1 Apple plus 1 apple

the time required to assemble computers varies directly as the number of computers assembled and inversely as the number of work

olga2289 [7]

Answer:

16 hours

Step-by-step explanation:

let t be time, n the number of computers and w the number of workers, then

t = $\frac{kn}{w}$ ← k is the constant of variation

To find k use the condition n = 30, w = 6 and t = 10 , then

10 = $\frac{30k}{6}$ ( multiply both sides by 6 )

60 = 30k ( divide both sides by 30 )

2 = k

t = $\frac{2n}{w}$ ← equation of variation

When w = 5 and n = 40 , then

t = $\frac{2(40)}{5}$ = $\frac{80}{5}$ = 16 hours

3 0

3 years ago

(8x − 5)(2x2 − 5x − 6)

Elodia [21]

Answer:

(-5x - 2)(8x - 5)

I hope this helps!

6 0

3 years ago

At the amusement park, Laura paid $6.00 for small cotton candy. Her older brother works at the park, and he told her they mark u

Rudiy27

$2 because 6/%300= 2

3 0

3 years ago

Read 2 more answers

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

There are 3 boxes the 1st Box contains the letters MATH the 2nd Box contains the letters AND the 3rd Box contains the letters HI

iogann1982 [59]

Answer:

2/14

Step-by-step explanation:

There are 14 letters in all of the boxes combined and only 2 of those letters are A.

8 0

3 years ago