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miss Akunina [59]

2 years ago

11

The points (–2, 4) and (–2, –2) are vertices of a heptagon. Explain how to find the length of the segment formed by these endpoi

nts. How long is the segment?

2 answers:

spin [16.1K]2 years ago

8 0

Answer:

Sample Response: The segment is vertical because the x-coordinates are the same. To find the distance, you can count the spaces between the points. The segment is 6 units long.

Step-by-step explanation:

Scrat [10]2 years ago

5 0

Answer:

The segment is 6 units long.

Step-by-step explanation:

The points (–2, 4) and (–2, –2) are vertices of a heptagon. We have to explain how to find the length of the segment formed by these endpoints.

If two points at the ends of a straight line PQ are P( $x_{1},y_{1}$ ) and Q( $x_{2},y_{2}$ ), then the length of the segment PQ will be given by the formula

$\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$

Now, in our case the two points are (-2,4) and (-2,-2) and the length of the segment will be

$\sqrt{(- 2 - (- 2))^{2} + (4 - (- 2))^{2}} = 6$ units. (Answer)

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nirvana33 [79]

Answer:

$p_v = 2*P(t_{n-2} > |t_{calc}|)= 0.91$

So on this case for the significance level assumed $\alpha=0.05$ we see that $p_v >\alpha$ so then we can conclude that the result is NOT significant. And we don't have enough evidence to reject the null hypothesis.

So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

Step-by-step explanation:

Let's suppose that we have the following linear model:

$y= \beta_o +\beta_1 X$

Where Y is the dependent variable and X the independent variable. $\beta_0$ represent the intercept and $\beta_1$ the slope.

In order to estimate the coefficients $\beta_0 ,\beta_1$ we can use least squares procedure.

If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:

Null Hypothesis: $\beta_1 = 0$

Alternative hypothesis: $\beta_1 \neq 0$

Or in other words we want to check is our slope is significant (X have an effect in the Y variable )

In order to conduct this test we are assuming the following conditions:

a) We have linear relationship between Y and X

b) We have the same probability distribution for the variable Y with the same deviation for each value of the independent variable

c) We assume that the Y values are independent and the distribution of Y is normal

The significance level assumed on this case is $\alpha=0.05$

The standard error for the slope is given by this formula:

$SE_{\beta_1}=\frac{\sqrt{\frac{\sum (y_i -\hat y_i)^2}{n-2}}}{\sqrt{\sum (X_i -\bar X)^2}}$

Th degrees of freedom for a linear regression is given by $df=n-2$ since we need to estimate the value for the slope and the intercept.

In order to test the hypothesis the statistic is given by:

$t=\frac{\hat \beta_1}{SE_{\beta_1}}$

The p value on this case would be given by:

$p_v = 2*P(t_{n-2} > |t_{calc}|)= 0.91$

So on this case for the significance level assumed $\alpha=0.05$ we see that $p_v >\alpha$ so then we can conclude that the result is NOT significant. And we don't have enough evidence to reject the null hypothesis.

So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Westkost [7]

Answer:

a) $\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b) $Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

c) $P(\bar X >31.25)=0.006=0.6\%$

d) $P(\bar X >28.25)=0.9997=99.97\%$

e) $P(28.25$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =30,\sigma =2)$

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b. What is the standard error of the mean time?

The standard error is given by this formula:

$Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >31.25)=1-P(\bar X$

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >28.25)=1-P(\bar X$

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

$P(28.25$

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