If BC = 4 and CD = 27, find AC.

Jack took a test and got 13 out of 17 points. He took a make-up test and got 12 out of 15 points.

blagie [28]

Answer:

4.6

Step-by-step explanation:

The answer is 4.6%, hope it helps! sorry if wrong!

3 0

3 years ago

River C is 400 miles longer than River D. If the sum of their lengths is 5 comma 520 miles, what is the length of each river?

Viefleur [7K]

If the sum of their lengths is 5,520 miles then River D is 2360 and River C is 3160

4 0

3 years ago

Sofia has four times as many dollars as her friend jessie has. together the two have $35. how much money do they each have?

garik1379 [7]

4J = s
J + s = J + 4j = 5j = 35

J = 7
s = 28

3 0

2 years ago

The math team has 11 boys, 10 girls, and 1 coach. What is the ratio of coach to students? Group of answer choices 21/1 11/1 1/11

laila [671]

Answer:

1/21

Step-by-step explanation:

11 boys + 10 girls = 21 students

We want the ratio of coaches to students

1 coach: 21 students

1/21

8 0

3 years ago

Read 2 more answers

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

2 years ago