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fredd [130]
2 years ago
12

Select all the expressions that are equivalent to -36x+54y−90.

Mathematics
2 answers:
CaHeK987 [17]2 years ago
7 0

Answer:

b) -18(2x -3y + 5)

d) 18(-2x + 3y - 5)

e) -2(18x - 27y + 45)

Step-by-step explanation:

a) -9( 4x - 6y - 10) = -36x + 54y + 90 not equal

b) -18(2x - 3y + 5) = -36x + 54y - 90 equal

c) -6(6x + 9y - 15) = -36x - 54y + 90 not equal

d) 18(-2x + 3y - 5) = -36x + 54y - 90 equal

e) -2(18x - 27y + 45) = -36x + 54y - 90 equal

f) 2(-18x + 54y - 90) = -36x + 108y - 180 not equal

Lorico [155]2 years ago
6 0

Answer:

A. -9(4x-6y-10). B. -18(2x-3y+5)

=-36x+54y+90. = -36x+54y-90

=18xy+90. =18xy-90

=108xy. =-72xy

C. -6(6x+9y-15)

=-36x-54+90

=-90xy +90

=0xy

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lesantik [10]

Answer:

a) 0.778

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c) 0.6826

d) 0.3174

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Step-by-step explanation:

Given:

Sample size, n = 5

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P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

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Answer:

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Step-by-step explanation:

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Answer:

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