Which equation, when graphed, has x-intercepts at (-1,0) and (-5,0) and a y-intercept at (0, -30)?

The cost of pool service includes labor and supplies. labor is 80% of the total cost. the cost of labor for a certain pool servi

laila [671]

Answer:

I know im late but the answer is 275 I just took the test on k-12

Step-by-step explanation:

3 0

2 years ago

Write an equation that represents the volume V as a function of the height h

erma4kov [3.2K]

Answer:

h=V/b

Step-by-step explanation:

5 0

3 years ago

A right rectangular prism has a length of 9, a width of 4, and a height of 5. How many unit cubes can fit in the prism?

Serjik [45]

Answer:

The answer is 54 cubes.

Step-by-step explanation:

5 0

3 years ago

A nationwide study of American homeowners revealed that 65% have one or more lawn mowers. A lawnequipment manufacturer, located

GenaCL600 [577]

Answer:

$z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589$

$p_v =P(z>1.589)=0.056$

If we compare the p value obtained with the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ 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 homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

Step-by-step explanation:

Data given and notation

n=497 represent the random sample taken

X=340 represent the homes in Omaha with one or more lawn mowers

$\hat p=\frac{340}{497}=0.684$ estimated proportion of homes in Omaha with one or more lawn mowers

$p_o=0.65$ is the value that we want to test

$\alpha$ represent the significance level

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 true proportion of homes in Omaha with one or more lawn mowers is higher than 0.65.:

Null hypothesis: $p\leq 0.65$

Alternative hypothesis: $p > 0.65$

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:

$z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589$

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:

$p_v =P(z>1.589)=0.056$

If we compare the p value obtained with the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ 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 homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

7 0

3 years ago

WHY ARE THERE NO TUTORS AVIABLE PLEASE HELP

arsen [322]

Answer:

25+1.50m=80

she can travel 36.7 m

Step-by-step explanation:

25+1.5m=80

1.5m=80-25

1.5m=55

m=55/1.5

m=36.6666=36.7

7 0

3 years ago