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3 years ago

The mean of data set A is 43.5 and the MAD is 3.7. The mean of data set B is 12.8 and the MAD is 4.1.

Mathematics

2 answers:

Artemon [7]3 years ago

3 0

Answer:

data set B!

Step-by-step explanation:

egoroff_w [7]3 years ago

3 0

Answer:

For questions one: its B

Step-by-step explanation:

since, Data set B shows a greater degree of variability because the distance between a data point(value) and the mean is 4.1. This also means that the data points in Set B are more spread out compared to the data points in Set A.

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A hummingbird’s heart beats about 1,250 beats

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Answer:0﹤Ⅹ﹤1250

Step-by-step explanation:

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3 years ago

Bailey spent $12.25 at the arcade on Saturday and $9.50 at the arcade on Sunday. How much money did Bailey spend anogether?

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Answer:

21.75

Step-by-step explanation:

i added 12.25 by 9.50

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3 years ago

Is the square root of 16.45 a rational number or irrational number?

trapecia [35]

Answer:

Irrational

Step-by-step explanation:

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3 years ago

Read 2 more answers

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

$f(x)=(x-2)^{2}+1$

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

Then, the minimum point of this function is (3,1) and it corresponds to (A)

46)

Let's use the same method here.

$g(x)=f(x+2)$

$g(x)=((x+2)-2)^{2}+1$

$g(x)=(x)^{2}+1$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

Then, the minimum point of this function is (0,1) and it corresponds to (C)

47)

Let's use the same method here.

$g(x)=f(x)+2$

$g(x)=(x-2)^{2}+1+2$

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

Then, the minimum point of this function is (2,3) and it corresponds to (B)

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

$g(x)=(x-2)^{2}-2$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}-2$

$g(2)=-2$

Then, the minimum point of this function is (2,-2) and it corresponds to (D)

I hope it helps you!

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3 years ago

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Fed [463]

Answer:

Step-by-step explanation:

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3 years ago