Y= 3x + 2 and y= 3x -3. These lines are... parallel perpendicular neither

n 2018, homes in East Baton Rouge (EBR) Parish sold for an average of $239,000. You take a random sample of homes in Ascension p

Olenka [21]

Answer:

Conclusion

There is no sufficient evidence to conclude that the mean of the home prices from Ascension parish is higher than the EBR mean

Step-by-step explanation:

From the question we are told that

The population mean for EBR is $\mu_ 1 = \$239,000$

The sample mean for Ascension parish is $\= x_2 = \$246,000$

The p-value is $p-value = 0.045$

The level of significance is $\alpha = 0.01$

The null hypothesis is $H_o : \mu_2 = \mu_1$

The alternative hypothesis is $H_a : \mu_2 > \mu_1$

Here $\mu_2$ is the population mean for Ascension parish

From the data given values we see that

$p-value > \alpha$

So we fail to reject the null hypothesis

So we conclude that there is no sufficient evidence to conclude that the mean of the home prices from Ascension parish is higher than the EBR mean

3 0

3 years ago

Please help. Thank you

Inessa [10]

Given:

$PQRS\sim TUVW$

In the given figure, PS=x, RS=35, UV=20, VW=25 and TW=15

To find:

The scale factor from PQRS to TUVW.

Solution:

We have,

$PQRS\sim TUVW$

We know that the corresponding sides of similar figures are proportional. The scale factor is the ratio of one side of image and corresponding side of preimage.

The scale factor is:

$k=\dfrac{VW}{RS}$

$k=\dfrac{25}{35}$

$k=\dfrac{5}{7}$

Therefore, the scale factor from PQRS to TUVW is $k=\dfrac{5}{7}$ .

6 0

3 years ago

Majesty Video Production Inc. wants the mean length of its advertisements to be 26 seconds. Assume the distribution of ad length

Paladinen [302]

Answer:

a) By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.

b) s = 0.44

c) 0.84% of the sample means will be greater than 27.05 seconds

d) 98.46% of the sample means will be greater than 25.05 seconds

e) 97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation(also called standard error) $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 26, \sigma = 2, n = 21, s = \frac{2}{\sqrt{21}} = 0.44$

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

By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.

b. What is the standard error of the mean time? (Round your answer to 2 decimal places)

$s = \frac{2}{\sqrt{21}} = 0.44$

c. What percent of the sample means will be greater than 27.05 seconds?

This is 1 subtracted by the pvalue of Z when X = 27.05. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{27.05 - 26}{0.44}$

$Z = 2.39$

$Z = 2.39$ has a pvalue of 0.9916

1 - 0.9916 = 0.0084

0.84% of the sample means will be greater than 27.05 seconds

d. What percent of the sample means will be greater than 25.05 seconds?

This is 1 subtracted by the pvalue of Z when X = 25.05. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{25.05 - 26}{0.44}$

$Z = -2.16$

$Z = -2.16$ has a pvalue of 0.0154

1 - 0.0154 = 0.9846

98.46% of the sample means will be greater than 25.05 seconds

e. What percent of the sample means will be greater than 25.05 but less than 27.05 seconds?"

This is the pvalue of Z when X = 27.05 subtracted by the pvalue of Z when X = 25.05.

X = 27.05

$Z = \frac{X - \mu}{s}$

$Z = \frac{27.05 - 26}{0.44}$

$Z = 2.39$

$Z = 2.39$ has a pvalue of 0.9916

X = 25.05

$Z = \frac{X - \mu}{s}$

$Z = \frac{25.05 - 26}{0.44}$

$Z = -2.16$

$Z = -2.16$ has a pvalue of 0.0154

0.9916 - 0.0154 = 0.9762

97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds

8 0

3 years ago

Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboa

KengaRu [80]

Answer:

a. 0.5 or 50%

b. 0.496 or 49.6%

c. 0.8185 or 81.85%

d. Yes, it would be just 0.0013 or 0.13% of probability to find a typist whose speed exceeds 105 wpm

e. 0.0252 or 2.52%

f. The qualifying speed would be 47.4 wpm.

Step-by-step explanation:

a. Let's find the z-score this way:

μ = 60 σ= 15

z-score = (x - μ)/σ

z-score = (60 - 60)/15

z-score = 0

Now, let's calculate the p value for z-score = 0, using the z-table:

p (z=0) = 0.5 or 50%

b. z-score = (x - μ)/σ

z-score = (59.9 - 60)/15

z-score = -0.01

Now, let's calculate the p value for z-score = -.0.01, using the z-table:

p (z = -0.01) = 0.496 or 49.6%

If the question were less or equal than 60, a and b would have the same answer. But in this case, the question is "less than 60 wpm".

c.

z-score = (x - μ)/σ

z-score = (45 - 60)/15

z-score = - 1

Now, let's calculate the p value for z-score = -1, using the z-table:

p (z = -1) = 0.1587

z-score = (x - μ)/σ

z-score = (90 - 60)/15

z-score = 2

Now, let's calculate the p value for z-score = 2, using the z-table:

p (z = 2) = 0.9772

In consequence,

p (-1 ≤ z ≤ 2) = 0.9772 - 0.1587 = 0.8185 or 81.85%

d. z-score = (x - μ)/σ

z-score = (105 - 60)/15

z-score = 3

Now, let's calculate the p value for z-score = 3, using the z-table:

p (z = 3) = 0.9987

In consequence,

p (z > 3) = 1 - 0.9987 = 0.0013

Yes, it would be just a 0.13% of probability to find a typist whose speed exceeds 105 wpm.

e. z-score = (x - μ)/σ

z-score = (75 - 60)/15

z-score = 1

Now, let's calculate the p value for z-score = 1, using the z-table:

p (z = 1) = 0.8413

In consequence,

p (z > 1) = 1 - 0.8413 = 0.1587

and if the two typists are independently selected, then

p = 0.1587 * 0.1587

p = 0.0252 or 2.52%

f. p = 0.2, using the z-table, the z-score is -0.84, then:

z-score = (x - μ)/σ

-0.84 = (x - 60)/15

-12.6 = x - 60

-x = -60 + 12.6

-x = - 47.4

x = 47.4

The qualifying speed would be 47.4 wpm.

8 0

3 years ago

HURRY PLEASE!! I WILL GIVE BRAINLIEST

user100 [1]

Answer:

1/64

Step-by-step explanation:

1/64=(1/8)^2

1/64=(1/4)^3

6 0

3 years ago

Read 2 more answers

Y= 3x + 2 and y= 3x -3. These lines are...​parallel perpendicularneither

Y= 3x + 2 and y= 3x -3. These lines are...parallel perpendicularneither