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jonny [76]
3 years ago
5

Exterior Angles and Isosceles Triangles

Mathematics
2 answers:
Naddika [18.5K]3 years ago
7 0
Exterior angles =180-75=105
Interior angles =180_69=111
Ad libitum [116K]3 years ago
3 0

Step-by-step explanation:

this is not an isoceles triangle.

anyway, we need to remember 2 things :

1. the sum of all angles in a triangle is always 180°.

2. the sum of all angles around a single point on a line is 360° (as this point wound be the center of an invisible circle). and as the line cuts this circle in half, the angles on one side of the line around this point is therefore 180°.

so, the internal angle Q is then

180 = 50 + 55 + Q

Q = 180 - 50 - 55 = 180 - 105 = 75°

the exterior angle x at Q is then

180 = Q + x = 75 + x

x = 180 - 75 = 105°

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GenaCL600 [577]
Hey there!

This linear equation is written in standard form, a form in which it's quite easy to find intercepts for one main reason. This reason revolves around zeros. When you see an x intercept, there's always a zero for the y, and when there's a y you see a zero for the x. Therefore, we can do this to get the intercepts.

For x:

2x-5(0) =10
2x = 10
x = (5,0)

For y:

2(0) -5y = 10
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y= (0,-2)

Now, to find the slope, we can easily convert it to slope intercept form:

y = mx+b

Where m is the slope and b is he y intercept.

2x-5y=10
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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.

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    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.

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Answer:

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Step-by-step explanation:

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