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3 years ago

15

There were a total of 9 soccer games during the three month season. If the games are equally divided, how many soccer games are

played a month

1 answer:

weqwewe [10]3 years ago

5 0

Answer: 3

Step-by-step explanation:

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The zeros for f(x)=x^4-4x^3-9x^2+36x are

Mrrafil [7]

X^4 - 4x^3 - 9x^2 + 36x = 0
x(x^3 - 4x^2 - 9x + 36) = 0
x(x^2(x - 4) - 9(x - 4) = 0
x(x + 3)(x - 3)(x - 4) = 0

the zeroes are -3, 0, 3, 4 Answer

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3 years ago

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Step-by-step explanation:

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What is the inverse of f(x)=5x-8

Ksju [112]

Answer:

$\large\boxed{f^{-1}(x)=\dfrac{1}{5}x+\dfrac{8}{5}}$

Step-by-step explanation:

$f(x)=5x-8\to y=5x-8\\\\\text{change}\ x\ \text{to}\ y\ \text{and vice versa}\\\\x=5y-8\\\\\text{solve for}\ y:\\\\5y-8=x\qquad\text{add 8 to both sides}\\\\5y-8+8=x+8\\\\5y=x+8\qquad\text{divide both sides by 5}\\\\\dfrac{5y}{5}=\dfrac{x}{5}+\dfrac{8}{5}\\\\y=\dfrac{1}{5}x+\dfrac{8}{5}$

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3 years ago

The probability that a can of paint contains contamination is 3.23%, and the probability of a mixing error is 2.4%. The probabil

stich3 [128]

Answer:

4.6%.

Step-by-step explanation:

The probability that a can of paint contains contamination(C) is 3.23%

P(C)=3.23%

The probability of a mixing(M) error is 2.4%.

P(M)=2.4%

The probability of both is 1.03%.

$P(C \cap M)=1.03\%$

We want to determine the probability that a randomly selected can has contamination or a mixing error. i.e. $P(C \cup M)$

In probability theory:

$P(C \cup M) = P(C)+P(M)-P(C \cap M)\\P(C \cup M)=3.23+2.4-1.03\\P(C \cup M)=4.6\%$

The probability that a randomly selected can has contamination or a mixing error is 4.6%.

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4 years ago