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2 years ago

Confusing confusion please help

Mathematics

2 answers:

Levart [38]2 years ago

8 0

Divide the shape into two parts, square triangle, the square is easy 2x2=4. The second part you know the length is 2 as 4-2=2, and because it is half of a square, it will be half of the area, so total answer is 6

White raven [17]2 years ago

5 0

Answer:

Step-by-step explanation:

Area = [(major base + minor base) x height]/2 = [(2+4)x2]/2 = 6

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Nissan wants to administer a satisfaction survey to its current customers. Using their customerâ€‹ database, the company randoml

Irina-Kira [14]

Answer: simple random sampling

Step-by-step explanation:

A simple random sampling is a sampling method in which individuals are chosen randomly from the population. In this each individual has equal chances to get selected.

If N is the population size , then for any individual , the chances for getting selected is $\dfrac{1}{N}$ .

As per given , the company randomly selects 50 customers directly from the database without applying any further steps and asks them about their level of satisfaction with the company.

Thus , the sampling technique used here is simple random sampling.

3 0

3 years ago

GIVING AWAY A LOT POINTS! PLEASE ANSWER!

masya89 [10]

The slope of the line is “rise over run.” That’s the vertical change between the two points (the difference in the y-coordinates) divided by the horizontal change over the same segment (the difference in the x-coordinates).

6 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

3 years ago

What is the range of g(x)?

Mila [183]

Answer:

Step-by-step explanation:

I just need for the points

5 0

2 years ago

What is the average rate of change of f over the interval [-1, 4] Give an exact number.

Sonbull [250]

Answer:

1.4

Step-by-step explanation:

The average rate of change is the "rise" divided by the "run".

rise/run = (f(4) -f(-1))/(4 -(-1)) = (0 -(-7))/(4+1)

rise/run = 7/5 = 1.4

The average rate of change on the interval [-1. 4] is 1.4.

3 0

2 years ago