(2, 4) and (6, 12) find slope

Pls help me with this i am giving 50pts out

podryga [215]

I think the answers is B

3 0

2 years ago

The surface area of a rectangular prism is 15 m2.

ch4aika [34]

$1.5$ × $10^5$ = 150,000

You answer is C. 150,000 cm²

4 0

3 years ago

8a – 10 = 6a in words

Ipatiy [6.2K]

Answer:

10 subtracted from the product of 8 and 'a' is equal to the product of 6 and 'a'

Step-by-step explanation:

3 0

3 years ago

What is the factor of the trinomial <img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" al

lara [203]

Refer to the attached for step by step explaination.

Have a nice day...

7 0

3 years ago

4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B

AleksAgata [21]

Answer:

$z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123$

$p_v =P(Z>2.123)=0.0169$

The p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .

Step-by-step explanation:

1) Data given and notation

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

$\hat p=\frac{63}{200}=0.315$ estimated proportion of lawyers had used some form of advertising for their business

$p_o=0.25$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that more than 25% of all the lawyers in Boston advertise for their business:

Null hypothesis: $p\leq 0.25$

Alternative hypothesis: $p > 0.25$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123$

4) 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:

$p_v =P(Z>2.123)=0.0169$

The p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .

8 0

3 years ago