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(x – 6)2 = y + 7, find the vertex.

Anna007 [38]

Would be the answer 16

3 0

3 years ago

QUESTION 2

PilotLPTM [1.2K]

Step-by-step explanation:

no of the above madam...

3 0

3 years ago

(Refer to the attached image)

gayaneshka [121]

Explanation:

Provided <u>2 length sides</u> and <u>one angle</u> also need to find <u>one missing side</u>.

So, use cosine rule:

a² = b² + c² - 2bc cos(A)

c² = 9² + 11² - 2(11)(9) cos(57)

c² = 94.16147

c = √94.16147 = 9.70 cm

d² = 5² + 7² - 2(5)(7) cos(48)

d² = 27.16

d = √27.16 = 5.21

5² = 7² + 9² - 2(7)(9) cos(H)

-126cos(H) = 25 - 49 - 81

cos(H) = -105/-126

cos(H) = 5/6

H = cos⁻¹(5/6) = 33.56°

8² = 4² + 7² - 2(4)(7) cos(J)

-56cos(J) = 64 - 16 - 49

cos(J) = -1/-56

J = cos⁻¹(1/56) = 88.98°

Passcode: 3142

8 0

2 years ago

Please help !!!!!!!!

Annette [7]

The answer is D cuz it cross the x

5 0

3 years ago

A group of retired admirals, generals, and other senior military leaders, recently published a report, "Too Fat to Fight". The r

weqwewe [10]

Answer:

$z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735$

$p_v =P(z$

If we compare the p value obtained and 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 americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .

Step-by-step explanation:

1) Data given and notation

n=180 represent the random sample taken

X=125 represent the number of americans between 17 to 24 that not qualify for the military

$\hat p=\frac{125}{180}=0.694$ estimated proportion of americans between 17 to 24 that not qualify for the military

$p_o=0.75$ 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 less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :

Null hypothesis: $p\geq 0.75$

Alternative hypothesis: $p < 0.75$

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:

$z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735$

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 left tailed test the p value would be:

$p_v =P(z$

If we compare the p value obtained and 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 americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .

6 0

3 years ago