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3 years ago

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QUESTION 6 What is the mean of the following set of data? Round your answer to two decimal places. 78, 82, 91, 100, 78, 92, 82,

77, 80
a. 91.21
b. 86.75
c. 84.44
d. 75.23

QUESTION 7

Find the median of the following set of data.
13, 11, 4, 5, 6, 9, 10, 12, 15, 16

a. no median

b. 10

c. 9

d. 10.5

2 answers:

guajiro [1.7K]3 years ago

8 0

Mean is the average number. To find mean, add all the numbers and divide by how many numbers there are.

$78 + 82 + 91 + 100 + 78 + 92 + 82 + 77 + 80 = 760$

There are 9 numbers. Divide by 9

$760 \div 9 = 84.44$

Question 6 ⇒ C. 84.44

Median is the middle number when ordered from least to greatest. First order the numbers.

$4, 5, 6, 9, 10, 11, 12, 13, 15, 16$

The 2 middle numbers are 10 and 11. Add them together and divide by 2

$10 + 11 = 21 \\ 21 \div 2 = 10.5$

Question 7 ⇒ D. 10.5

crimeas [40]3 years ago

7 0

C and D are the answers for 6 and 7. hope this helps:)

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Consider ΔABC, whose vertices are A (2, 1), B (3, 3), and C (1, 6); let the line segment AC represent the base of the triangle.

(a) Find the equation of the line passing through B and perpendicular to the line AC

(b) Let the point of intersection of line AC with the line you found in part A be point D. Find the coordinates of point D.

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Step-by-step explanation:

Given

$\triangle ABC$

$A = (2,1)$

$B = (3,3)$

$C = (1,6)$

Solving (a): Line that passes through B, perpendicular to AC.

First, calculate the slope of AC

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$C = (1,6)$ --- $(x_2,y_2)$

The slope is:

$m = \frac{6- 1}{1 - 2}$

$m = \frac{5}{-1}$

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The slope of the line that passes through B is calculated as:

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So, we have:

$m_2 = -\frac{1}{-5}$

$m_2 = \frac{1}{5}$

The equation of the line is the calculated using:

$m_2 = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$m_2 = \frac{1}{5}$

$B = (3,3)$ --- $(x_1,y_1)$

$(x_2,y_2) = (x,y)$

So, we have:

$\frac{1}{5} = \frac{y - 3}{x - 3}$

Cross multiply

$5(y-3) = 1(x - 3)$

$5y - 15 = x - 3$

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Make y the subject

$y = \frac{1}{5}x + \frac{12}{5}$

Solving (b): Point of intersection between AC and $y = \frac{1}{5}x + \frac{12}{5}$

First, calculate the equation of AC using:

$y = m(x - x_1) + y_1$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$m = -5$

So:

$y=-5(x - 2) + 1$

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$y=-5x + 11$

So, we have:

$y=-5x + 11$ and $y = \frac{1}{5}x + \frac{12}{5}$

Equate both to solve for x

i.e.

$y = y$

$-5x + 11 = \frac{1}{5}x + \frac{12}{5}$

Collect like terms

$-5x -\frac{1}{5}x = \frac{12}{5} - 11$

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$x = \frac{43}{26}$

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$y = -5 * \frac{43}{26} + 11$

$y = \frac{-5 *43}{26} + 11$

Take LCM

$y = \frac{-5 *43+11 * 26}{26}$

$y = \frac{71}{26}$

Hence, the coordinates of D is:

$D = (\frac{43}{26},\frac{71}{26})$

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