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3 years ago

11

How do you write 1.7 in expanded form

2 answers:

olga2289 [7]3 years ago

7 0

<span>1.7 = (1 x 1) + (7/10)

Hope this helps! Could I gent a Brainliest?
</span>

sergij07 [2.7K]3 years ago

4 0

The answer would be 7/10

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What is the simplest form of 7/10 + 1/5

earnstyle [38]

Solutions

1) Find the least common denominator of 7/10 and 1/5.
10: 10
5 : 5 , 10
The least common denominator is 10.

2) Make the denominator of 1/5 to 10. Multiply numerator and denominator by 2.

1 x 2 = 2
5 x 2 = 10

3) New fraction
7/10 + 2/10

4) Add
7+2 = 9

9/10 is answer

9/10 can not be simplified. It is already in reduced form.

7 0

3 years ago

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If 20 squirrels can eat a bushel of corn in 10 hours, how many squirrels can eat a bushel of corn in 2 hours?

omeli [17]

So if 20 squirrels can eat a bushel of corn in 10 hours that means that 2 it needs 2 squirrels per hour.

So the answer would be 4 squirrels, hope this helps!

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3 years ago

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Llana [10]

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2 years ago

What is the cube of the quotient 24 and 6

Naily [24]

Quotient of 24 and 6 is : $\frac { 24 }{ 6 } =4$

and 4's cube is : ${ 4 }^{ 3 }\quad =\quad 4\cdot 4\cdot 4=64$
The answer is 64

7 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 $\frac{2}{11}$ .

4 0

2 years ago