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slava [35]
3 years ago
7

The means and mean absolute deviations of the amount of rain that fell each day in a local city, last week and this week, are sh

own below. Means and Mean Absolute Deviations of Rainfall Last Week and This Week Last Week This Week Mean 3.5 in. 2.7 in. Mean Absolute Deviation 1.2 in. 0.5 in. Which expression compares the difference of the two means to this week's mean absolute deviation?
Mathematics
2 answers:
jekas [21]3 years ago
8 0

Answer:

0.8 / 0.5

Step-by-step explanation:

Given :

Deviations of Rainfall :

______Last Week ____ This Week

Mean___ 3.5 in. ________2.7 in

MAD ___ 1.2 in _________ 0.5 in

Mean difference = 3.5 in - 2.7 in = 0.8 in

This week's mean absolute deviation = 0.5 in

Mean difference / this week's mean absolute deviation

= 0.8 / 0.5

Wittaler [7]3 years ago
7 0

Answer:

0.8 / 0.5

Step-by-step explanation:

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Find the distance between the points (-7, -6) and (-3, 0).
FrozenT [24]

Answer:

7.21 unit

Step-by-step explanation:

distance= √[(-7+3)^2 + (-6-0)^2]

= 7.21 unit

5 0
3 years ago
(06.07)
Zinaida [17]
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3. If the scatter plot's points loosely scattered going down to the right, this is a negative slope, and thus it is a negative correlation.
4. We would expect the data to be negatively correlated, since the longer a person has been jogging, the more tired he would be.
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6 0
4 years ago
Read 2 more answers
The difference of 8 and 5?
Bess [88]

Answer:

3

Step-by-step explanation:

8 - 5 = 3

3 0
3 years ago
Read 2 more answers
What is the value of 1/2+1/3
RideAnS [48]

Answer:

5/6

Step-by-step explanation:

1/2=3/6

1/3=2/6

______

5/6

4 0
3 years ago
hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and
Mariulka [41]

Given:

The equation is,

2\log _3x-\log _3(x-2)=2

Explanation:

Simplify the equation by using logarthimic property.

\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{      \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}

Simplify further.

\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}

Solve the quadratic equation for x.

\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

\begin{gathered} x-6=0 \\ x=6 \end{gathered}

For (x - 3) = 0,

\begin{gathered} x-3=0 \\ x=3 \end{gathered}

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0
1 year ago
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