What is the distance between the points (-1, -7 ) and (4, 5)

Keisha needs to make at least 32 costumes for the school play. Since she can make 4 costumes each week, Keisha plans to work on

Licemer1 [7]

Complete question is;

Keisha needs to make at least 32 costumes for the school play. Since she can make 4 costumes each week, Keisha plans to work on the costumes for at least 8 weeks.

Which inequality matches the problem?

A) x ≥ 8

B) x = 8

C) x ≤ 8

D) none of the above

Answer:

A: x ≥ 8

Step-by-step explanation:

Let the number of weeks be x.

She needs to make at least 32 costumes and plans to work at least 8 weeks. Since she makes 4 costumes per week, we can say that;

4x ≥ 32

I used greater than or equal to sign because she can't make less than 32 costumes.

Also, she can't use less than 8 weeks.

Thus; dividing both sides by 4, we have;

4x/4 ≥ 32/4

x ≥ 8

4 0

3 years ago

In a recent study, researchers found that 31 out of 150 boys aged 7-13 were overweight or obese. On the basis of this study can

AURORKA [14]

Answer:

1) Null hypothesis: $p \leq 0.15$

Alternative hypothesis: $p > 0.15$

2) 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) $z_{crit}= 1.64$

And the rejection zone would be $z>1.64$

4) Calculate the statistic

$z=\frac{0.207 -0.15}{\sqrt{\frac{0.15(1-0.15)}{150}}}=1.955$

5) Statistical decision

For this case our calculated value is on the rejection zone, so we have enough evidence to reject the null hypothesis at 5% of significance and we can conclude that the true proportion is higher than 0.15

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=21 represent the boys overweight

$\hat p=\frac{31}{150}=0.207$ estimated proportion of boy overweigth

$p_o=0.15$ 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)

1) Concepts and formulas to use

We need to conduct a hypothesis in order to test the true proportion of boys obese is higher than 0.15.:

Null hypothesis: $p \leq 0.15$

Alternative hypothesis: $p > 0.15$

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)

2) 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) Decision rule

For this case we need a value on the normal standard distribution who accumulates 0.05 of the area on the right tail and on this case this value is:

$z_{crit}= 1.64$

And the rejection zone would be $z>1.64$

4) Calculate the statistic

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

$z=\frac{0.207 -0.15}{\sqrt{\frac{0.15(1-0.15)}{150}}}=1.955$

5) Statistical decision

For this case our calculated value is on the rejection zone, so we have enough evidence to reject the null hypothesis at 5% of significance and we can conclude that the true proportion is higher than 0.15

6 0

3 years ago

A certain intelligence test has an N(100, 100) distribution. To see whether intelligence is inherited, tests are given to the el

Anettt [7]

Answer:

Follows are the solution to the given choices:

Step-by-step explanation:

In choice a:

$\to \mu = 100$