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3 years ago

Which of the following situations most likely results in a negative correlation when the data is displayed on a scatter plot?

2 answers:

4 0

It's B because the car gets older as the years of ownership go up; resulting in a negative correlation on a graph. Hope I helped :)

Nataliya [291]3 years ago

3 0

Answer:

Negative correlation is determined by: Option: b

b) the age of car vs. the value of the car

Step-by-step explanation:

We know that a negative correlation is one in which the increase in one variable leads to the decrease ion the other variable i.e. the two variables have a inverse relation.

the # of hours vs. the # of miles a person travels

This is an example of a positive correlation because with the increasing number of hours while travelling the more distance a person will travel.

Hence, option (a) is incorrect.

the age of car vs. the value of the car

This is an example of a negative correlation.

Because as the commodity gets older it's price also get reduced.

c) the # of years spent in school vs. salary earned

There is no correlation between the two variables as the time spent in school is nowhere related to salary.

Hence, option (c) is incorrect.

d) the age of a house vs. the amount of people who have lived in a house.

Here with the increase of one variable the other variable may increase or decrease it can't necessarily decrease.

Hence, option (d) is incorrect.

Hence, the correct option is:

b) the age of car vs. the value of the car

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-3 1/5 ÷ 1 3/4<br> Write your answer as a mixed number in simplest form.

Ipatiy [6.2K]

So the answer would be negative 64 over 35

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3 years ago

Which point on the number line best represents the location of 78?

Sliva [168]

Answer:

Can you show the whole question

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3 years ago

Read 2 more answers

Consider this equation.

mina [271]

Answer:

No, Mia is not correct, the answer is -2.

Step-by-step explanation:

First we are going to use the distributive Property.

Step 1: 2(0.75+0.4). multiply the 2 out. .75m x 2=1.5m and 2 x .4= .8.

We get 1.5m+.8, so 2(0.75+0.4)=1.5m+.8

Now we are going to distribute on the oher side. 4(.5m-.8)

4x .5m=2m and 4x.8=3.2. Keep the sign the same and we get 2m-3.2, so 4(.5m-.8)=2m-3.2

We put this into the new equation to get 1.5m+.8+7.8=-6.4m+2m-3.2

Step 2: combine like terms

.8+7.8= 8.6 and -6.4m +2m =-4.4m

Plug that into our equation and we get 1.5mn+8.6=-4.4m-3.2

Step 3: Switch the sides

Bring -4.4m to the other side to get 1.5m+4.4m=5.9m

One side is 5.9m

We bring 8.6 to the other side to get -8.6. now we do -3.2-8.6=-11.8.

Our new equation is 5.9m=-11.8

divide by 5.9 on both sides to get our answer of

m=-2

6 0

3 years ago

Read 2 more answers

2 people working at the same rate will take 6 hours to paint a room.

inna [77]

Answer:

4 hours

Step-by-step explanation:

2×6=3x

12÷3= 4

x=4 so 4 hours.

8 0

2 years ago

Read 2 more answers

Grades on a standardized test are known to have a mean of 1000 for students in the United States. The test is administered to 45

vovikov84 [41]

Answer:

a. The 95% confidence interval is 1,022.94559 < μ < 1,003.0544

b. There is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. i. The 95% confidence interval for the change in average test score is; -18.955390 < μ₁ - μ₂ < 6.955390

ii. There are no statistical significant evidence that the prep course helped

d. i. The 95% confidence interval for the change in average test scores is 3.47467 < μ₁ - μ₂ < 14.52533

ii. There is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

Step-by-step explanation:

The mean of the standardized test = 1,000

The number of students test to which the test is administered = 453 students

The mean score of the sample of students, $\bar{x}$ = 1013

The standard deviation of the sample, s = 108

a. The 95% confidence interval is given as follows;

$CI=\bar{x}\pm z\dfrac{s}{\sqrt{n}}$

At 95% confidence level, z = 1.96, therefore, we have;

$CI=1013\pm 1.96 \times \dfrac{108}{\sqrt{453}}$

Therefore, we have;

1,022.94559 < μ < 1,003.0544

b. From the 95% confidence interval of the mean, there is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. The parameters of the students taking the test are;

The number of students, n = 503

The number of hours preparation the students are given, t = 3 hours

The average test score of the student, $\bar{x}$ = 1019

The number of test scores of the student, s = 95

At 95% confidence level, z = 1.96, therefore, we have;

The confidence interval, C.I., for the difference in mean is given as follows;

$C.I. = \left (\bar{x}_{1}- \bar{x}_{2} \right )\pm z_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

Therefore, we have;

$C.I. = \left (1013- 1019 \right )\pm 1.96 \times \sqrt{\dfrac{108^{2}}{453}+\dfrac{95^{2}}{503}}$

Which gives;

-18.955390 < μ₁ - μ₂ < 6.955390

ii. Given that one of the limit is negative while the other is positive, there are no statistical significant evidence that the prep course helped

d. The given parameters are;

The number of students taking the test = The original 453 students

The average change in the test scores, $\bar{x}_{1}- \bar{x}_{2}$ = 9 points

The standard deviation of the change, Δs = 60 points

Therefore, we have;

C.I. = $\bar{x}_{1}- \bar{x}_{2}$ + 1.96 × Δs/√n

∴ C.I. = 9 ± 1.96 × 60/√(453)

i. The 95% confidence interval, C.I. = 3.47467 < μ₁ - μ₂ < 14.52533

ii. Given that both values, the minimum and the maximum limit are positive, therefore, there is no zero (0) within the confidence interval of the difference in of the means of the results therefore, there is statistically significant evidence that students will perform better on their second attempt after the prep course

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2 years ago