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4 years ago

Please help me with all thank you in advance

Mathematics

2 answers:

balu736 [363]4 years ago

8 0

Answer:

1232i34iu494232759375r

Step-by-step explanation:

ser-zykov [4K]4 years ago

8 0

1) 0

2) 2/6

3) 2/6

4) 8/10

5) 2/10

6) 1/6

7) 1/4

8) 6/10

9) 7/12

10) 3/10

11) 4/5

12) 2/5

13) 3/3=1

14) 5/12

15) 5/8

16) 11/12

17) 1(2/4)

18) 1(1/4)

19) 1(1/10)

20) 1(3/10)

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Please answer the question from the attachment.

VashaNatasha [74]

32 divided by (3+7) = 3.2
3.2x3= 9.6
3.2x7= 22.4
Therefore 9.6:22.4

8 0

3 years ago

Read 2 more answers

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salantis [7]

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3 years ago

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madam [21]

What is the question that you're asking?

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4 years ago

543,000,000,000,000 in Scientific notation

Zielflug [23.3K]

Answer:

5.43×10^14 or

$5 .43 \times 10^{14}$

Step-by-step explanation:

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3 years ago

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,

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

5 0

3 years ago