Is the Mean Value Theorem applicable to the function f(x) = |x - 1| on the interval [0, 2]?Why or why not?

Answers

Answer 1

Answer: The only point that derivative of the function f(x) = |x - 1| is not continuous is at x = 0. You need to check whether the slope for the interval (0,2) is continuous to see if you can apply MVT. The interval (0,2) does not include end points, so 0 is not in this interval. The function is continuous over the interval, so MVT can be applied.

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Help Me! I don't understand this problem!

Answers

Answer: A) -0.83

Without the actual numeric data coordinates, it is impossible to compute the true r value. Though we can estimate. The data points are negatively correlated in a fairly strong manner. We can draw a straight line close to all of these points, so the r value is going to be fairly close to -1. The closer r is to -1, the stronger the negative correlation. Having r = -1 exactly means all of the points fall on some single straight line that slopes downward.

Choice B is the next best choice, but its correlation isn't as strong. So that's why I ruled it out. Choices C and D are ruled out immediately since they are positive values.

Whats 9+10? dont answer wrong

Answers

Step-by-step explanation:

when we add 9 and 10 we will get sum 19

Answer:

Step-by-step explanation:

meme

Solve for the value of v.

Answers

Answer:

6x9 =54

54-4 =50

v =9

Answer:

v =9

Step-by-step explanation:

Which of the following equations are the correct answer? Please help due tomorrow

Answers

(3,2), (-1,-4)

Slope = (y2-y1)/(x2-x1)= (2+4)/(3+1) =3/2
(y-y1) = 3/2(x-x1)
(y-2)=3/2(x-3)
(y+4)=3/2(x+1)
Correct answer: A,E

A.3(f) The line graphed on the grid represents the first of two equations in a system of linear equations.20
-16
12
-8
-20 -16 -12
-
-4
48
12
16 2024
-8
-12
16
If the graph of the second equation in the system passes through (-12, 20) and (4,12), which statement is true?

Answers

I don’t any idea for this

You want to create an 80% confidence interval for the average age of people who attend U of O football games. You take a sample of 100 attendees and find the average age to be 43.7 years old with a standard deviation of 7 years. Find the value of t* for this confidence interval. Do not round your answer. Write your answer in decimal form, not as a fraction or percent.

Answers

Answer:

80% confidence interval for the average age of people who attend U of O football games is [42.795 , 44.604].

Step-by-step explanation:

We are given that a sample of 100 attendees and find the average age to be 43.7 years old with a standard deviation of 7 years.

So, the pivotal quantity for 80% confidence interval for the population average start up cost is given by;

P.Q. = $(\bar X - \mu)/((s)/(√(n) ) )$ ~ $t_n_-_1$

where, $\mu$ = sample average age = 43.7 years old

$\sigma$ = sample standard deviation = 7 years

n = sample of attendees = 100

$\mu$ = population average age of people

So, 80% confidence interval for the average age of people, $\mu$ is ;

P(-1.2915 < $t_9_9$ < 1.2915) = 0.80

P(-1.2915 < $(\bar X - \mu)/((s)/(√(n) ) )$ < 1.2915) = 0.80

P( $-1.2915 * {(s)/(√(n) )$ < ${\bar X - \mu}$ < $1.2915 * {(s)/(√(n) )$ ) = 0.80

P( $\bar X -1.2915 * {(s)/(√(n) )$ < $\mu$ < $\bar X +1.2915 * {(s)/(√(n) )$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X -1.2915 * {(s)/(√(n) )$ , $\bar X +1.2915 * {(s)/(√(n) )$ ]

= [ $43.7 -1.2915 * {(7)/(√(100) )$ , $43.7 +1.2915 * {(7)/(√(100) )$ ]

= [42.795 , 44.604]

Therefore, 80% confidence interval for the population average age of people who attend U of O football games is [42.795 , 44.604].

Other Questions

To calculate approximately the amount or value of something is called _________.

If ST=19 and S lies at -4 , where could T be located?

Please answer this correctly

The number of new bicycles must be more than 13 times the number of new treadmills. Each bicycle costs $340 and each treadmill costs $670. He must spend less than $5,650. Select all of the constraints for this situation. a) y>13x b)340x+670y≤5650 c)y>0 340x+670y<5650 d) x>13y x>0