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3 years ago

PLEASE HELP ILL GIVE 40 POINTS PLEASE ILL MARK BRAINLIST

Mathematics

1 answer:

Len [333]3 years ago

5 0

Answer:

The answer is G because the triangles are similar and the rise over run is the same for both.

Step-by-step explanation:

Hope this helps!

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4/5 is 0.8 into a fraction

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What is the y-intercept?

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-3.5

Step-by-step explanation:

Whenever the x is 0, the graph hits the y axis

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4 years ago

An inequality is shown below. −8x ≤ 38 Which inequality represents the solution set?

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3 years ago

3x3-27x=0 Solve by factoring a

Eddi Din [679]

After factoring $3x^3-27x=0$ the solutions are x=0 , x=3, x= -3

Step-by-step explanation:

We need to solve by factoring:

$3x^3-27x=0$

Solving:

$3x^3-27x=0$

Taking 3x common:

$3x(x^2-9)=0$

Now using formula $a^2-b^2=(a-b)(a+b)$

$3x((x)^2-(3)^2)=0\\3x(x-3)(x+3)=0$

$3x=0\,\,and\,\,x+3=0\,\,and\,\,x-3=0\\x=0\,\,and\,\,x=-3\,\,and\,\,x=3\\$

So, After factoring $3x^3-27x=0$ the solutions are x=0 , x=3, x= -3

Keywords: Factorization

Learn more about Factorization at:

#learnwithBrainly

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3 years ago

Clarkson University surveyed alumni to learn more about what they think of Clarkson. One part of the survey asked respondents to

erastova [34]

Answer:

(a) The probability that a randomly selected alumnus would say their experience surpassed expectations is 0.05.

(b) The probability that a randomly selected alumnus would say their experience met or surpassed expectations is 0.67.

Step-by-step explanation:

Let's denote the events as follows:

A = Fell short of expectations

B = Met expectations

C = Surpassed expectations

N = no response

Given:

P (N) = 0.04

P (A) = 0.26

P (B) = 0.65

(a)

Compute the probability that a randomly selected alumnus would say their experience surpassed expectations as follows:

$P(C) = 1 - [P(A) + P(B) + P(N)]\\= 1 - [0.26 + 0.65 + 0.04]\\= 1 - 0.95\\= 0.05$

Thus, the probability that a randomly selected alumnus would say their experience surpassed expectations is 0.05.

(b)

The response of all individuals are independent.

Compute the probability that a randomly selected alumnus would say their experience met or surpassed expectations as follows:

$P(B\cup C) = P(B)+P(C)-P(B\cap C)\\=P(B)+P(C)-P(B)\times P(C)\\= 0.65 + 0.05 - (0.65\times0.05)\\=0.6675\\\approx0.67$

Thus, the probability that a randomly selected alumnus would say their experience met or surpassed expectations is 0.67.

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4 years ago