Which of the following graphs represents the line of slope 3/2 that passes through the point (1,

An insurance agent says the standard deviation of the total hospital charges for patients involved in a crash in which the vehic

lisov135 [29]

Let the null hypothesis be that the agent's claim is correct. Using a standard normal probability table we find:
$p-value=P(Z \ \textgreater \ 1.1)=0.14$
The p-value 0.14 is much larger than alpha = 0.01, so we cannot reject the agent's claim.

7 0

3 years ago

Bárbara corre con rapidez constante de 5 km/h a lo largo de 500 m que separan el colegio de su casa. Demora 3 minutos en recoger

sergiy2304 [10]

Answer:

try socratic it should help

6 0

3 years ago

In a geometry class, the students were asked to prove the theorem below by contradiction.

Readme [11.4K]

<span>The contradiction term for "at most one obtuse angle" is "no obtuse angles
so therefore i conclude that the </span> statement she will most likely use as an assumption is
Let each angle of a triangle be acute
so option A is correct
hope it helps

4 0

3 years ago

Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer

agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming: the function is $f(x)=x^{2}$ in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

$g'(0)=g'(1)$

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

$g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0$

$g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2$

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

$f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]$

This is what the Bilateral Theorem says:

$\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L$

4 0

4 years ago

I need an answer not a link that don't work!

Mila [183]

Answer:

B

Step-by-step explanation:

6 0

3 years ago

Which of the following graphs represents the line of slope 3/2 that passes through the point (1, –1)?