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3 years ago

9

Direction: Choose the letter of the correct answer and encircle the letter that corresponds to it.

1 answer:

Mamont248 [21]3 years ago

7 0

Answer:

1) D =65

2) C = 52

3) D =18

4) B =10

5) A = 72 and 108

Step-by-step explanation:

1) D =65

2) C = 52

3) D =18

4) B =10

5) A = 72 and 108

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2. Suppose (220)_b and (251)_b are the squares of consecutive integers. Determine b.

Andrei [34K]

Let $x$ and $x+1$ be these consecutive integers. For some base $b$ , we can write

$(x_{10})^2=220_b=2b^2+2b$

$((x+1)_{10})^2=251_b=2b^2+5b+1$

If

$x^2=2b^2+2b$

then

$x^2+1=2b^2+2b+1$

Now,

$x^2+2x+1=2b^2+5b+1$

$2x+2b^2+2b+1=2b^2+5b+1$

$\implies2x=3b$

Squaring both sides gives

$4x^2=9b^2$

and so

$4(2b^2+2b)=9b^2$

$\implies8b^2+8b=9b^2$

$\implies b^2-8b=0\implies\boxed{b=8}$

(because the base has to be non-zero)

3 0

3 years ago

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 $\frac{2}{11}$ .

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3 years ago

Yesterday, Grace drove 2 1/8 miles. she used 1 1/4

prisoha [69]

Answer:

7/8

Step-by-step explanation:

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3 years ago

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Select the relationship that does represent a function.

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Answer:

c

Step-by-step explanation:

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3 years ago

if it is friday then letica family will have pizza for dinner. if letica family has pizza for dinner, letica will song at the di

stiv31 [10]

Answer:

the first statement

Step-by-step explanation:

it's the first one because on the second one the family members name is letica and the family is letica so it cant be both

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3 years ago