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2 years ago

I’m confused on this one

Mathematics

1 answer:

Mandarinka [93]2 years ago

3 0

Remark

If the lines are parallel then triangle RQS will be similar to triangle RTP

From that, all three lines in one triangle will bear the same ratio to all three lines of the second triangle.

Givens

PQ = 8

QR = 5

RS = 15

ST = x + 3

Ratio

QR/RP = RS/RT

Sub and solve

RP = 5 + 8

RP = 13

RT = 15 + x + 3

RT = 18 + x

5/13 = 15 / (18 + x) Cross multiply

5(18 + x) = 195 Remove the brackets on the left.

90 + 5x = 195 Subtract 90 from both sides.

5x = 105 Divide by 5

x = 105/5

x = 21 Answer <<<<<<<

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Answer:

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Step-by-step explanation:

1) d+3< 1

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Answer:

1. $P(s_1|I)=\frac{1}{11}$

2. $P(s_2|I)=\frac{8}{11}$

3. $P(s_3|I)=\frac{2}{11}$

Step-by-step explanation:

Given information:

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$P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1$

(1)

We need to find the value of P(s₁|I).

$P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

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Therefore the value of P(s₁|I) is $\frac{1}{11}$ .

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We need to find the value of P(s₂|I).

$P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}$

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$P(s_2|I)=\frac{8}{11}$

Therefore the value of P(s₂|I) is $\frac{8}{11}$ .

(3)

We need to find the value of P(s₃|I).

$P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}$

$P(s_3|I)=\frac{0.03}{0.165}$

$P(s_3|I)=\frac{2}{11}$

Therefore the value of P(s₃|I) is $\frac{2}{11}$ .

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