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Helen [10]
3 years ago
11

PLEASE HELP!!!!! WORTH 50 POINTS! PLEASE ANSWER ALL QUESTIONS (6,7,8,9)

Mathematics
1 answer:
iogann1982 [59]3 years ago
7 0

Answer:

6. Additional information is required on the two triangles

7. The pair of triangles are similar by SAS similarity postulate

8. Yes, by triangle proportionality theorem

9. DE = 9

Step-by-step explanation:

When two or more triangles are similar, the ratio of the corresponding sides of both triangles is constant

6. From the given figure, more information is required on the two triangles

However, whereby the side with length 10 is parallel to the corresponding side on the other triangle, then the alternate angles formed by the transversal to the two parallel lines are equal and the two triangles are similar by Angle Angle, AA, similarity postulate

7. From the given triangles, the ratio of the corresponding sides relative to the given included angles are;

18/45 and 10/25

18/45 = 2/5

10/25 = 2/5

Therefore, given that two pair of corresponding sides have the same included angle and that the two included corresponding sides of the of the two triangles are proportional, then the two triangles are similar by the Side Angle Side SAS similarity postulate

8. By triangle proportionality theorem, we have;

If \overline{MN} ║ \overline{KL}, then we have;

3/6 = 4/8 = 2

∴  \overline{MN} ║ \overline{KL}

9. By triangle proportionality theorem, where \overline{BE} ║ \overline{CD}, we have;

12/20 = DE/15

∴ DE = 15 × 12/20 = 9

DE = 9

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\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

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Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

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We can calculate each point of the joint probability as:

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X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3

X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05

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X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025

X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1

X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18

X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03

X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015

X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015

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X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12

X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02

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X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01

X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04

We can write it in the form of a matrix:

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56

We can calculate P(X<= 1) * P(Y<=1)

P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56

Both calculations give the same result.

c) Probability of no violations

P(X+Y=0)=P(0,0)=0.3

d) P(X + Y <= 1)

P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53

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