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3 years ago

Which shows the graph of f(x)= |x|-2

Mathematics

1 answer:

elena55 [62]3 years ago

6 0

The answer is C), because it represents the vertical translation of the graph. It shifts 2 units down due to -2

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What is the equation of the line that passes through (0,-3) and (2, 2)?

tigry1 [53]

Answer:

$y = \frac{5}{2}x -3$

Step-by-step explanation:

Given points (0,-3) (2, 2), we can start by solving for the slope of the line by using the formula:

$m = \frac{y2 - y1}{x2 - x1}$

Let (x1, y1) = (0,-3)

(x2, y2) = (2, 2)

Plug in these values into the slope formula:

$m = \frac{y2 - y1}{x2 - x1} = \frac{2 - (-3)}{2 - 0} = \frac{5}{2}$

Therfore, the slope of the line is $\frac{5}{2}$ .

Next, we must determine the y-intercept of the line. By definition, the y-intercept is the y-coordinate of the point where the graph of the linear equation crosses the y-axis. The y-intercept is also the value of y when x = 0.

Using the slope (m) = $\frac{5}{2}$ and one of the given points, (2, 2), plug in these values into the slope-intercept form, y = mx + b:

y = mx + b

$y = \frac{5}{2}x + b$

$2 = \frac{5}{2}(2) + b$

2 = 5 + b

Subtract 5 on both sides of the equation to solve for b:

2 - 5 = 5 + b - 5

-3 = b

The y-intercept (b) = -3.

Therefore, the equation of the line that passes through (0,-3) and (2, 2) is:

$y = \frac{5}{2}x -3$ .

4 0

3 years ago

A supplier of a certain car parts chain of stores wants to estimate the average length of time car owners plan to keep their car

____ [38]

Answer:

The 99% confidence interval for the average length of time all car owners plan to keep their cars is between 3.85 years and 10.55 years.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575*\frac{6.5}{\sqrt{25}} = 3.35$

The lower end of the interval is the sample mean subtracted by M. So it is 7.2 - 3.35 = 3.85 years

The upper end of the interval is the sample mean added to M. So it is 7.2 + 3.35 = 10.55 years

The 99% confidence interval for the average length of time all car owners plan to keep their cars is between 3.85 years and 10.55 years.

7 0

3 years ago

26 times does go into 92

sleet_krkn [62]

Well i am doing my math on my paper and really its a long answer3.53846153846

6 0

3 years ago

If the last digit of a number is 0 or 5, then the number is divisible by 10. True False

Whitepunk [10]

Answer:

False

Step-by-step explanation:

15 isn't divisible by 10

3 0

3 years ago

Read 2 more answers

Assuming that the population is normally distributed, construct a 9090% confidence interval for the population mean for each o

jasenka [17]

Given:

Set A: 1 4 4 4 5 5 5 8

Mean: 4.5

Standard dev: 1.9

Set B:

Mean: 4.5

Standard dev: 2.45

% = 90

Set A:

Standard Error, SE = s/ √n = 1.9/√8 = 0.67

Degrees of freedom = n - 1 = 8 -1 = 7

t- score = 1.89457861

Width of the confidence interval = t * SE = 1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

Set B:

Standard Error, SE = s/ √n = 2.45/√8 = 0.87

Degrees of freedom = n - 1 = 8 -1 = 7

t- Score = 1.89457861

Width of the confidence interval = t * SE = 1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width = 4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width = 4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.

6 0

4 years ago