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3 years ago

Which equation has a graph that is a parabola with a vertex at (–1, –1)?

Mathematics

2 answers:

Artist 52 [7]3 years ago

3 0

Answer:

$y=(x+1)^2-1$

which agrees with the last option listed.

Step-by-step explanation:

We use the vertex form of a parabola:

$y-y_v=a(x-x_v)^2$

which with the information we have becomes:

$y-y_v=a(x-x_v)^2\\y-(-1)_v=a(x-(-1))^2\\y=a(x+1)^2-1$

Therefore the equation that corresponds to such vertex (-1, -1) is the last one listed among your given answer options.

$y=(x+1)^2-1$

Sophie [7]3 years ago

3 0

Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

Kevin Sims bought a car for $12,000. A year is was worth 9,000. What was the percent decrease in value?

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3 years ago

Consider the following hypothesis test:

postnew [5]

Answer:

a. P-value = 0.039.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. P-value = 0.013.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. P-value = 0.130.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean significantly differs from 100.

Then, the null and alternative hypothesis are:

$H_0: \mu=100\\\\H_a:\mu\neq 100$

The significance level is 0.05.

The sample has a size n=65.

The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

a. The sample mean is M=103.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.5.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.5}{\sqrt{65}}=1.4264$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{103-100}{1.4264}=\dfrac{3}{1.4264}=2.103$

This test is a two-tailed test, with 64 degrees of freedom and t=2.103, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>2.103)=0.039$

As the P-value (0.039) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. The sample mean is M=96.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11}{\sqrt{65}}=1.3644$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{96.5-100}{1.3644}=\dfrac{-3.5}{1.3644}=-2.565$

This test is a two-tailed test, with 64 degrees of freedom and t=-2.565, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t$

As the P-value (0.013) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. The sample mean is M=102.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=10.5.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{10.5}{\sqrt{65}}=1.3024$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{102-100}{1.3024}=\dfrac{2}{1.3024}=1.536$

This test is a two-tailed test, with 64 degrees of freedom and t=1.536, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>1.536)=0.130$

As the P-value (0.13) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

8 0

3 years ago

PLZZZZZZZZ HELPPPP!!!!!!!!!!!!!!

levacccp [35]

Answer:

1 135

2 80

3 145

4 45

Step-by-step explanation:

Thats the answer

7 0

3 years ago