Mathematics, 27.03.2021 18:30 pamelakroll98

Brian recorded the number of commercials played during two types of television shows: an educational documentary and a primetime drama. Both types of shows lasted 1 hour, including commercials. He recorded the number of commercials for 10 episodes of each type of show.

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Answer from: cmay454

m(x)_A = 5.2 --- Mean absolute deviation of educational documentary

m(x)_B = 4.7 --- Mean absolute deviation of prime time drama

Step-by-step explanation:

Given

n = 10

See attachment for table

Required

Determine the mean absolute deviation of each

Mean absolute deviation m(x) is calculated as:

$m(x) = \frac{1}{n}\sum |x - \mu|$

For Educational Documentary

First, calculate the mean

$\mu = \frac{\sum x}{n}$

So:

$\mu_A = \frac{26 + 28 + 30 + 18+27 + 18 + 20 +31+17+17}{10}$

$\mu_A = \frac{232}{10}$

$\mu_A = 23.2$

The mean absolute deviation is then calculated as:

$m(x)_A = \frac{1}{n}\sum |x - \mu_A|$

$m(x)_A = \frac{1}{10}(|26 -23.2| +|28 -23.2| +|30 -23.2| +|18 -23.2| +|27 -23.2| +|18 -23.2| +|20 -23.2| +|31 -23.2| +|17 -23.2| +|17 -23.2|)$

$m(x)_A = \frac{1}{10}(|2.8| +|4.8| +|6.8| +|-5.2| +|3.8| +|-5.2| +|-3.2| +|7.8| +|-6.2| +|-6.2|)$

$m(x)_A = \frac{1}{10}(2.8 +4.8 +6.8 +5.2 +3.8 +5.2 +3.2 +7.8 +6.2 +6.2)$

$m(x)_A = \frac{1}{10}*52$

m(x)_A = 5.2

For Prime time Drama

First, calculate the mean

$\mu = \frac{\sum x}{n}$

So:

$\mu_B = \frac{39+39+35+29+40+27+41+29+32+30}{10}$

$\mu_B = \frac{341}{10}$

$\mu_B = 34.1$

The mean absolute deviation is then calculated as:

$m(x)_B = \frac{1}{n}\sum |x - \mu_B|$

$m(x)_B = \frac{1}{10}(|39-34.1|+|39-34.1|+|35-34.1|+|29-34.1|+|40-34.1|+|27-34.1|+|41-34.1|+|29-34.1|+|32-34.1|+|30-34.1|)$

$m(x)_B = \frac{1}{10}(|4.9|+|4.9|+|0.9|+|-5.1|+|5.9|+|-7.1|+|6.9|+|-5.1|+|-2.1|+|-4.1|)$

$m(x)_B = \frac{1}{10}(4.9+4.9+0.9+5.1+5.9+7.1+6.9+5.1+2.1+4.1)$

$m(x)_B = \frac{1}{10}*47$

m(x)_B = 4.7

Brian recorded the number of commercials played during two types of television shows: an educational

Answer from: Quest

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

$ab^2+ac^2=bc^2$

$ab^2+ac^2=x^2$ )

triangle abd and cba are similar,

$\frac{ab}{bc}=\frac{bd}{ab}$

$ab^2=bc\cdot bd$

$25=bc\cdot bd$ (2)

triangle cad and cba are similar,

$\frac{ac}{bc}=\frac{cd}{ac}$

$ac^2=bc\cdot dc$

$7\times 7=bc\cdot dc$

$49=bc\cdot dc$ )

using (1), (2) and (3), we get

$25+49=x^2$

$74=x^2$

hence proved.

therefore option 2 is correct.

Answer from: Quest

a. we have two

lines: y = 4-x and

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b. 4-x = 8-x^-1

x 4-x 8-x^-1

-3 7

8.33

-2 6 8.5

-1 5 9

0 4

1 3

2 2

7.5

3 1 7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:

x=-4.236

y = 4-x = 4 – (-4.236)

= 8.236

y = 8-x^-1 = .236)^-1

= 8.236

x=0.236

y = 4-x = 4 – (0.236) = 3.764

y = 8-x^-1 = 8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c. to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x and y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines. the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

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