Question 4

eimsori [14]

Couple things to note:

Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.
Slope can be calculated using any two points on a line and the formula y₁ - y₂ / x₁ - x₂.

For the first problem, we know the slope of Function A is 6 (refer to slope-intercept form above). To compare the slopes of Function A and Function B, first find the slope of Function B.

Use y₁ - y₂ / x₁ - x₂. Two points on the line are (0, 1) and (-1, -2). Plug these into the formula accordingly and solve for slope.

y₁ - y₂ / x₁ - x₂

1 - (-2) / 0 - (-1)

1 + 2 / 0 + 1

3 / 1

3

The slope of Function B is 3. This is half of 6 (the slope of Function A), so the correct answer to question 1 is the first option: Slope of Function B = 2 × Slope of Function A.

For the second problem, substitute m and b in y = mx + b according to the graph. b is the y-intercept (the point at which the line intersects the y-axis); it is (0, -4), or -4. This gives us

y = mx - 4

We must now find m. Follow the same steps above to find slope. Our two points are (-2, 0) and (0, -4).

y₁ - y₂ / x₁ - x₂

0 - (-4) / -2 - 0

0 + 4 / -2

4 / -2

-2

Substitute.

y = -2x - 4

The first option is the correct answer.

5 0

3 years ago

Read 2 more answers

over the course of 4 days the price of stock at ACME corporation fell $28. What was the average change in stock price per day?

expeople1 [14]

28/4 = 7...and since the price fell, it would be -7 per day

8 0

3 years ago

Read 2 more answers

The numbers 1 through 50 are written with red marker on slips of paper while the numbers 51 through 100 are written with blue ma

Marizza181 [45]

Given:
1 - 50 written in red marker
51 - 100 written in blue marker

Probability of selecting a number greater than 81; 100 - 81 = 19 possible numbers. 1 draw. 1/19

Probability of selecting a number written in red: 1/50

Probability of selecting a number written in blue: 1/50

Probability of selecting a number that is a multiple of 10. There are 10 instances; 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 ; 1/10

5 0

3 years ago

Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas

polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 120-1 = 119$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have $T = 1.6578$ .

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{650}{\sqrt{120}} = 59.34$

Now, we multiply T and s

$M = T*s = 59.34*1.6578 = 98.37$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.95$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

With 119 and 0.025 in the t-distribution table, we have $T = 1.9801$ .

$M = T*s = 59.34*1.9801 = 117.50$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that $\alpha = 0.99$

So

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005$

With 119 and 0.025 in the t-distribution table, we have $T = 2.6178$ .

$M = T*s = 59.34*2.6178 = 155.34$

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0

3 years ago

Solve fort: 7 alg 2NE AS Stuck? Watch a video or hisobi Gggggb zigzag

kozerog [31]

Wat does that mean
Confused!!!

7 0

3 years ago