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BARSIC [14]
3 years ago
10

How would I solve n^2=2n

Mathematics
1 answer:
lidiya [134]3 years ago
5 0

Answer: n=0 or n=2

Step-by-step explanation: Step 1: Subtract 2n from both sides.

n²−2n=2n−2n

n²−2n=0

Step 2: Factor left side of equation.

n(n−2)=0

Step 3: Set factors equal to 0.

n=0 or n−2=0

n=0 or n=2

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Which of the following represents the most accurate estimation of 65 + 77
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3 years ago
6 high school seniors choose from among 20 quotes for their yearbook. What is the probability that at least 2 of them choose the
shusha [124]

Using the binomial distribution, it is found that there is a 0.0328 = 3.28% probability that at least 2 of them choose the same quote.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem, we have that:

  • There are 6 students, hence n = 6.
  • There are 20 quotes, hence the probability of each being chosen is p = 1/20 = 0.05.

The probability of one quote being chosen at least two times is given by:

P(X \geq 2) = 1 - P(X < 2)

In which:

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

Then:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{6,0}.(0.05)^{0}.(0.95)^{6} = 0.7351

P(X = 1) = C_{6,1}.(0.05)^{1}.(0.95)^{5} = 0.2321

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.7351 + 0.2321 = 0.9672.

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9672 = 0.0328

0.0328 = 3.28% probability that at least 2 of them choose the same quote.

More can be learned about the binomial distribution at brainly.com/question/24863377

6 0
2 years ago
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