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3 years ago

Find the coordinate that divides the directed line segment from A(-2,-4) to B(8,1) in the ratio of 2 to 3

Mathematics

1 answer:

zalisa [80]3 years ago

7 0

We have that
A(-2,-4) B(8,1) 

let
M-------> the coordinate that divides the directed line segment from A to B in the ratio of 2 to 3

we know that

A--------------M----------------------B
2 3
distance AM is equal to (2/5) AB
distance MB is equal to (3/5) AB
so

step 1
find the x coordinate of point M
Mx=Ax+(2/5)*dABx
where
Mx is the x coordinate of point M
Ax is the x coordinate of point A
dABx is the distance AB in the x coordinate
Ax=-2
dABx=(8+2)=10
Mx=-2+(2/5)*10-----> Mx=2

step 2
find the y coordinate of point M
My=Ay+(2/5)*dABy
where
My is the y coordinate of point M
Ay is the y coordinate of point A
dABy is the distance AB in the y coordinate
Ay=-4
dABy=(1+4)=5
Mx=-4+(2/5)*5-----> My=-2

the coordinates of point M is (2,-2)

see the attached figure

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Read 2 more answers

You roll 2 dice. If the sum of the dice is: 2, 3, 4 you win $20. If the sum is 5, 6, 7, 8 you win $10. If the sum is 9, 10, 11 y

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1) The expected value of the "winnings" is $(-0.97 $\overline 2$ )

2) The variance for the "winnings" is $0.57966

3) The standard deviation for the "winnings" is$0.761354

4) The game is not a fair game because one is expected to lose $0.97 $\overline 2$

Step-by-step explanation:

1) The probability of having a sum of 2 = 1/6×1/6 = 1/36

The probability of having a sum of 3 = 1/6×1/6 = 1/36

The probability of having a sum of 4 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 5 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 6 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12

The probability of having a sum of 7 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12

The probability of having a sum of 8 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12

The probability of having a sum of 9 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 10 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 11 = 1/6×1/6 = 1/36

The probability of having a sum of 12 = 1/6×1/6 = 1/36

The values are;

For 2, we have 1/36 × (20 - 5) = 0.41 $\overline 6$

For 3, we have 1/36 × (20 - 5) = 0.41 $\overline 6$

For 4, we have 1/18 × (20 - 5) = 0.8 $\overline 3$

For 5, we have 1/18 × (10 - 5) = 0.2 $\overline 7$

For 6, we have 1/12 × (10 - 5) = 0.41 $\overline 6$

For 7, we have 1/12 × (10 - 5) = 0.41 $\overline 6$

For 8, we have 1/12 × (10 - 5) = 0.41 $\overline 6$

For 9, we have 1/18 × (-20 - 5) = -1.3 $\overline 8$

For 10, we have 1/18 × (-20 - 5) = -1.3 $\overline 8$

For 11, we have 1/36 × (-20 - 5) = -0.69 $\overline 4$

For 12, we have 1/36 × (-25 - 5) = -0.69 $\overline 4$

The expected value of the winnings is given as follows;

0.41 $\overline 6$ + 0.41 $\overline 6$ + 0.8 $\overline 3$ + 0.8 $\overline 3$ + 0.8 $\overline 3$ + 0.41 $\overline 6$ + 0.41 $\overline 6$ + -1.3 $\overline 8$ -1.3 - 0.69 $\overline 4$ - 0.69 $\overline 4$ = -0.97 $\overline 2$

Therefore, the expected value = $-0.97 $\overline 2$ , which is one is expected to lose $0.97 $\overline 2$

2) Using Microsoft Excel, we have;

The variance for the "winnings", σ² = $0.57966

3) The standard deviation for the "winnings" = √σ² = √(0.57966) ≈ $0.761354

4) The game is not a fair game because one is expected to lose $0.97 $\overline 2$

3 0

3 years ago