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Oksana_A [137]
3 years ago
5

A pet store has 7 cats. Here are their weights (in pounds).

Mathematics
1 answer:
Alexandra [31]3 years ago
6 0
I believe the mean is 60 pounds
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Given an equation x3 - 8x2 - 9x + 72 = 0, find the zeros.
sveta [45]

Answer:

= 28/3

Step-by-step explanation:

Let's solve your equation step-by-step.

x(3)−(8)(2)−9x+72=0

Step 1: Simplify both sides of the equation.

x(3)−(8)(2)−9x+72=0

3x+−16+−9x+72=0

(3x+−9x)+(−16+72)=0(Combine Like Terms)

−6x+56=0

−6x+56=0

Step 2: Subtract 56 from both sides.

−6x+56−56=0−56

−6x=−56

Step 3: Divide both sides by -6.

−6x

−6

=

−56

−6

x=

28

3

Answer:

x=

28

3

7 0
3 years ago
Question 4 of 5
belka [17]

Answer:

Angle B is a straight angel. Angels a and c are bot 45 degrees.

Step-by-step explanation:

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3 years ago
What is the mean of the data set rounded to the nearest tenth 0.7 1.1 0.8 1.6 1.6 2.2 1.1
Rina8888 [55]
The mean (average) can be found by adding up all ur numbers, then dividing by how many numbers there are.

(0.7+1.1+0.8+1.6+1.6+2.2+1.1) / 7 = 9.1 / 7 = 1.3 <==


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3 years ago
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Solve for x. Help me out because I haven’t been paying attention all year pt.2
Strike441 [17]

Answer:

x = 9

Step-by-step explanation:

(whole secant) x (external part) = (tangent)^2

(24+3) * 3 = x^2

27*3 = x^2

81 = x^2

Take the square root of each side

sqrt(81) = sqrt(x^2)

9 =x

3 0
3 years ago
The time it takes for a planet to complete its orbit around a particular star is called the? planet's sidereal year. The siderea
BartSMP [9]

Answer:

(a) See below

(b) r = 0.9879  

(c) y = -12.629 + 0.0654x

(d) See below

(e) No.

Step-by-step explanation:

(a) Plot the data

I used Excel to plot your data and got the graph in Fig 1 below.

(b) Correlation coefficient

One formula for the correlation coefficient is  

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}

The calculation is not difficult, but it is tedious.

(i) Calculate the intermediate numbers

We can display them in a table.

<u>    x   </u>    <u>      y     </u>   <u>       xy     </u>    <u>              x²    </u>   <u>       y²    </u>

   36       0.22              7.92               1296           0.05

   67        0.62            42.21              4489           0.40

   93         1.00            93.00           20164           3.46

 433        11.8          5699.4          233289        139.24

 887      29.3         25989.1          786769       858.49

1785      82.0        146370          3186225      6724

2797     163.0         455911         7823209    26569

<u>3675 </u>  <u> 248.0  </u>    <u>   911400      </u>  <u>13505625</u>   <u> 61504        </u>

9965   537.81     1545776.75  25569715   95799.63

(ii) Calculate the correlation coefficient

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}

(c) Regression line

The equation for the regression line is

y = a + bx where

a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy  - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} -  \dfrac{9\times 1545776.75  - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is $\large \boxed{\mathbf{y = -12.629 + 0.0654x}}$}

(d) Residuals

Insert the values of x into the regression equation to get the estimated values of y.

Then take the difference between the actual and estimated values to get the residuals.

<u>    x    </u>   <u>      y     </u>   <u>Estimated</u>   <u>Residual </u>

    36        0.22        -10                 10

    67        0.62          -8                  9

    93        1.00           -7                  8

   142        1.86           -3                  5

  433       11.8             19               -  7

  887     29.3             45               -16  

 1785     82.0            104              -22

2797    163.0            170               -  7

3675   248.0            228               20

(e) Suitability of regression line

A linear model would have the residuals scattered randomly above and below a horizontal line.

Instead, they appear to lie along a parabola (Fig. 2).

This suggests that linear regression is not a good model for the data.

4 0
3 years ago
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