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3 years ago

12

Okay so I think I got it right, I just need someone to check over it.

2 answers:

Paladinen [302]3 years ago

6 0

Answer:

yes

Step-by-step explanation:

alexira [117]3 years ago

4 0

Answer:

Yes, from what I learned its correct.

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Factor 4x^4+8x^2+1<br> PLZ try hurry ;-;.....anyone?

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2 years ago

Please help! 15 points!

MA_775_DIABLO [31]

Answers:

(a) Area: 2108 in² Perimeter: 192 in

(b) Area

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Step-by-step explanation:

To find the area, you do 34 × 62

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2 years ago

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Step-by-step explanation:

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2 years ago

8 + 4f = 3f solve for f

navik [9.2K]

Answer:

$f=-8$

Step-by-step explanation:

$8+4f=3f$

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Hope this helps!

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3 years ago

Read 2 more answers

Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl

amm1812

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:

$P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3$

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

(1)

We need to find the value of P(s₁|I).

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

$P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}$

$P(s_1|I)=\frac{0.015}{0.165}$

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

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

(2)

We need to find the value of P(s₂|I).

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

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

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

$P(s_2|I)=\frac{0.12}{0.165}$

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

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 $\frac{2}{11}$ .

4 0

2 years ago