What is the length of A'B'?

The following table shows the percent increase of donations made on behalf of a non-profit organization for the period of 1984 t

pashok25 [27]

Giving the table below which shows the percent increase of donations made on behalf of a non-profit organization for the period of 1984 to 2003.

Year: 1984 1989 1993 1997 2001 2003

Percent: 7.8 16.3 26.2 38.9 49.2 62.1

The scatter plot of the data is attached with the x-axis representing the number of years after 1980 and the y-axis representing the percent increase of donations made on behalf of a non-profit organization.

To find the equation for the line of regression where the x-axis representing the number of years after 1980 and the y-axis representing the percent increase of donations made on behalf of a non-profit organization.
$\begin{center}
\begin{tabular}{ c| c| c| c| }
 x & y & x^2 & xy \\ [1ex] 
 4 & 7.8 & 16 & 31.2 \\ 
 9 & 16.3 & 81 & 146.7 \\ 
13 & 26.2 & 169 & 340.6 \\ 
17 & 38.9 & 289 & 661.3 \\ 
21 & 49.2 & 441 & 1,033.2 \\ 
23 & 62.1 & 529 & 1,428.3 \\ [1ex]
\Sigma x=87 & \Sigma y=200.5 & \Sigma x^2=1,525 & \Sigma xy=3,641.3 
\end{tabular}
\end{center}$

Recall that the equation of the regression line is given by
$y=a+bx$
where
$a= \frac{(\Sigma y)(\Sigma x^2)-(\Sigma x)(\Sigma xy)}{n(\Sigma x^2)-(\Sigma x)^2} = \frac{200.5(1,525)-87(3,641.3)}{6(1,525)-(87)^2} \\ \\ = \frac{305,762.5-316793.1}{9,150-7,569} = \frac{-11,030.6}{1,581} =-6.977$
and
$b= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{n(\Sigma x^2)-(\Sigma x)^2} = \frac{6(3,641.3)-(87)(200.5)}{6(1,525)-(87)^2} \\ \\ = \frac{21,847.8-17,443.5}{9,150-7,569} = \frac{4,404.3}{1,581} =2.7858$

Thus, the equation of the regresson line is given by
$y=-6.977+2.7858x$

The graph of the regression line is attached.

Using the equation, we can predict the percent donated in the year 2015. Recall that 2015 is 35 years after 1980. Thus x = 35.

The percent donated in the year 2015 is given by
$-6.977+2.7858(35)=-6.977+97.503=90.526$

Therefore, the percent donated in the year 2015 is predicted to be 90.5

7 0

3 years ago

Two blocks are connected by a very light string passing over a massless and frictionless pulley. The 20.0 N block moves 75.0cm t

Angelina_Jolie [31]

The string is assumed to be massless so the tension is the sting above the 12.0 N block has the same magnitude to the horizontal tension pulling to the right of the 20.0 N block. Thus,
1.22 a = 12.0 - T (eqn 1)
and for the 20.0 N block:
2.04 a = T - 20.0 x 0.325 (using µ(k) for the coefficient of friction)
2.04 a = T - 6.5 (eqn 2)

[eqn 1] + [eqn 2] → 3.26 a = 5.5
a = 1.69 m/s²

Subs a = 1.69 into [eqn 2] → 2.04 x 1.69 = T - 6.5
T = 9.95 N

Now want the resultant force acting on the 20.0 N block:
Resultant force acting on the 20.0 N block = 9.95 - 20.0 x 0.325 = 3.45 N
Units have to be consistent ... so have to convert 75.0 cm to m: 

75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m
work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J

3 0

2 years ago

Rosa's best time in a race is two seconds faster than her second-best time. Her second-best time is 1 1/4 times her best time. W

Svetradugi [14.3K]

B) 10 seconds. 10/8=1 1/4

8 0

2 years ago

Find the area of the triangle with ∠A = 86, b = 4 ft, and c = 3 ft. Round your answer to two decimal places.

Leona [35]

I hope this helps you

Area=1/2.b.c.sinA

Area =1/2.4.3.0,99

Area =5,98

6 0

2 years ago

Campbell Middle school has 912 students.How many computers are in this schools computer lab?

Aleksandr [31]

You can't really determine the amount of computers from this question, they're obviously not going to have one computer for everyone so half third, or even a fourth of the school population

3 0

2 years ago

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