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lozanna [386]
3 years ago
10

Two months ago, the mean daily rainfall in a local city was 9.4 cm. The mean absolute deviation was 3.5 cm. Last month, the mean

daily rainfall in that city was 11.5 cm, and the mean absolute deviation was 1.6 cm.
Which statement about the rainfall is true?

More rain fell two months ago than last month.

About the same amount of rain fell two months ago as last month.

Last month, the amount of rain that fell each day varied about the same as the month before.

Last month, the amount of rain that fell each day varied less than the month before.
Mathematics
2 answers:
vova2212 [387]3 years ago
8 0

The actual answer is D. The amount of rain that fell each day varied less than the month before.


I just took the quiz

tino4ka555 [31]3 years ago
5 0
The statement that is true about rain fall is not Last month, the amount of rain that fell each day varied about the same as the month before. The answer in this question is not Last month, the amount of rain that fell each day varied about the same as the month before.
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The opposite angles of a rhombus are equal. <br> a. True <br> b. False
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Answer:

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Step-by-step explanation:

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It took my question down twice here's a real math question I guess
aliina [53]

Answer:

n-3

Interval notation: (-\infty, -10)\cup(-3,\infty)

Step-by-step explanation:

<u>First inequality:</u>

<u />n+8

Therefore, this inequality restricts:

n \in \mathrm{R};\: n

<u>Second inequality:</u>

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Therefore, this inequality restricts:

n \in \mathrm{R};\: n>-3

Therefore, with both of these restrictions together, we have:

\fbox{$n \in \mathrm{R}; n-3$}\\\mathrm{or\:}\fbox{$n-3$}\\\mathrm{or\:}\fbox{$(-\infty, -10)\cup(-3,\infty)$}.

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scZoUnD [109]

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Step-by-step explanation:

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3 years ago
The graph of a quadratic function passes through the points (-1,0), (-2,0), and (0,2). Write the quadratic function in general f
ddd [48]

Answer:

f(x)=x^2+3x+2

Step-by-step explanation:

step 1

we have the points

(-1,0), (-2,0), and (0,2)

Plot the points

using a graphing tool

see the attached figure

The graph of a quadratic function must be a vertical parabola open upward

The vertex is a minimum

The quadratic function in general form is equal to

f(x)=ax^2+bx+c

Substitute the value of x and the value of y of each given ordered pair in the general equation and solve for a,b and c

(0,2)

For x=0, y=2

substitute

2=a(0)^2+b(0)+c

c=2

(-1,0)

For x=-1, y=0

substitute

0=a(-1)^2+b(-1)+2

0=a-b+2 ----> a=b-2 ----> equation A

(-2,0)

For x=-2, y=0

substitute

0=a(-2)^2+b(-2)+2

0=4a-2b+2 ----> equation B

we have the system

a=b-2 ----> equation A

0=4a-2b+2 ----> equation B

substitute equation A in equation B

0=4(b-2)-2b+2

solve for b

0=4b-8-2b+2

2b=6

b=3

Find the value of a

a=3-2=1

therefore

The quadratic function in general form is equal to

f(x)=x^2+3x+2

see the attached figure N 2 to better understand the problem

7 0
3 years ago
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