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2 years ago

Rewrite the equation below in slope-intercept form. 2y−8x=18

Mathematics

2 answers:

zavuch27 [327]2 years ago

5 0

2y - 8x = 18

2y = 18 + 8x

y = 8x/2 + 18/2

y = 4x + 9

Natali5045456 [20]2 years ago

3 0

Reformat the equation so that it is in the form y=

2y=8x+18
y=4x+9

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Given q(x)=3x and r(x)=x2. What is (r∘q)(2)? 36 10 12 24

zmey [24]

Answer:

the answer would be 36.

Step-by-step explanation:

and here go you would get it

R(-5) = -3*(-5) - 4

R(-5) = 15 - 4

R(-5) = 11

Q(X)=3X+3

Q(11) = 3*11 + 3

Q(11) = 33 + 3

Q(11) = 36

So, (Q(R(-2) = 36

hope it is correct..

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Which of the following mortgage options will not have a PMI requirement?

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I think the answer c

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balandron [24]

I hope this helps you

Euclid

h^2=4.6

h=2 square root of 6

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24+36=base^2

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3 years ago

Solve for x. 3x - 1 = 27

poizon [28]

X = 9
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After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni

antoniya [11.8K]

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

$\hat p=\frac{64}{100}=0.64$ estimated proportion for were in favor of firing the coach

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

$0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719$

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =100*0.64=64>10$

$n(1-p_o)=100*(1-0.64)=36>10$

Calculate the statistic

The statistic is calculated with the following formula:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$

On this case the value of $p_o=0.5$ is the value that we are testing and n = 100.

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8$

The p value for the test would be:

$p_v =P(z>2.8)=1-P(z$

Part c

Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

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3 years ago