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2 years ago

Find the GCF from the two numbers, and rewrite the sum using the distributive property. 18-12=.

Mathematics

1 answer:

natulia [17]2 years ago

3 0

Answer:

ab+ac=a(b+c)

where a is the greatest common factor

find gcf of each

factor each

18=2*3*3

12=2*2*3

GCF=2*3=6

18+12=6(3)+6(2)=6(3+2)

Step-by-step explanation:

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Bezzdna [24]

Answer:

3.75

Step-by-step explanation:

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7 0

2 years ago

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asambeis [7]

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2 years ago

ΔABC is dilated by a scale factor of 0.25 with the origin as the center of dilation, resulting in the image ΔA′B′C′. If A = (-1,

AfilCa [17]

Answer:

c. d = 5

Step-by-step explanation:

The difference in coordinates B and C is (14, 17) -(2, 1) = (12, 16). When the figure is scaled by 0.25, that difference becomes ...

0.25(12, 16) = (3, 4)

The length of a line segment with those coordinate differences is ...

d = √(3² +4²) = √25 = 5 . . . . matches selection c.

3 0

2 years ago

The probability of winning the game is 7/10. What is the probability that they will lose?

cestrela7 [59]

3/10 is the probability of them losing.

8 0

3 years ago

Suppose that X has a Poisson distribution with a mean of 64. Approximate the following probabilities. Round the answers to 4 dec

o-na [289]

Answer:

(a) The probability of the event (X > 84) is 0.007.

(b) The probability of the event (X < 64) is 0.483.

Step-by-step explanation:

The random variable X follows a Poisson distribution with parameter λ = 64.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, ...$

(a)

Compute the probability of the event (X > 84) as follows:

P (X > 84) = 1 - P (X ≤ 84)

$=1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007$

Thus, the probability of the event (X > 84) is 0.007.

(b)

Compute the probability of the event (X < 64) as follows:

P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)

$=\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483$

Thus, the probability of the event (X < 64) is 0.483.

5 0

3 years ago