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3 years ago

12. I don’t know how to do it

Mathematics

1 answer:

Nostrana [21]3 years ago

8 0

Answer:

BC = 50

Step-by-step explanation:

I believe what I'm doing is correct but to be completely honest with you, I haven't done this in a while lol

Okay so...

You would do

1. 6y-16 = 4y+6

You would subtract 4 from 6...

2. 2y-16 = 6

You would then add 16 to 6...

3. 2y = 22

Then, divide 2 from 22...

4. y = 11

Substitute into BC...

5. 6 (11) - 16 = 66 - 16 = 50

Then, you get your final answer...

6. BC = 50

I hope it's right. Good luck ^^

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Answer:

The given statement is false.

Step-by-step explanation:

Reason

let D be a directed graph with 'n' no of vertices and 'E' edges.

where 'n'=1. thus D =(n,E).

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3 years ago

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Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

100 is the exterior angle.

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Formula : -

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2 years ago

Read 2 more answers

Suppose that 10% of all homeowners in an earthquake-prone area of California are insured against earthquake damage. Four homeown

guapka [62]

Remainder of question:

Find the probability distribution of x

Answer:

The random variable x is defined as: X = {0, 1, 2, 3, 4}

The probability distribution of X:

P(X = 0) = 0.656

P(X = 1) = 0.2916

P(X= 2) = 0.0486

P(X=3) = 0.0036

P(X = 4) = 0.0001

Step-by-step explanation:

Sample size, n = 4

Random variable, X = {0, 1, 2, 3, 4}

10% (0.1) of the homeowners are insured against earthquake, p = 0.1

Proportion of homeowners who are not insured against earthquake, q = 1 - 0.1

q = 0.9

Probability distribution of x,

$P(X = r) = ^nC_r *p^r q^{n-r} \\\\P(X= 0) =(^4C_0 *p^1 q^4 )\\P(X=0) = (^4C_0 *0.1^0 0.9^4 ) = 0.656\\P(X= 1)= (^4C_1 *p^1 q^3 )\\P(X=1) = (^4C_1 *0.1^1 0.9^3 ) = 0.2916\\P(X= 2)=( ^4C_2 *p^2 q^2) \\P(X=2) = (^4C_2 *0.1^2 0.9^2 ) = 0.0486\\P(X= 3) = (^4C_3 *p^3 q^3) \\ P(X=3) = (^4C_3 *0.1^3 0.9^1 ) = 0.0036\\P(X= 4) = (^4C_4 *p^4 q^0 )\\ P(X=4) =(^4C_4 *0.1^4 0.9^0 ) = 0.0001$

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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klio [65]

Answer:

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Step-by-step explanation:

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3 years ago