Wanting someone to help me with the proceduresn²+n=120

1 answer:

4 0

N^2 + n - 120 = 0

Quadratic formula
a = 1, b = 1, c = -120
n = [-1 plus or minus root(1^2 - 4(1)(-120))] / 2(1)
n = (-1 plus or minus root481)/2
n =(approximately) 10.5 or -11.5

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goblinko [34]

Answer:

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An electronic product contains 40 integrated circuits. The probability that any integrated circuit is defective is 0.01, and the

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Answer:

There is a 33.10% probability that there is at least one defective integrated circuit.

Step-by-step explanation:

For either integrated circuit, there are only two possible outcomes. Either they are defective, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And $\pi$ is the probability of X happening.

In this problem

The electonic product contains 40 integrated circuits, so $n = 40$ .

The probability that any integrated circuit is defective is 0.01, so $\pi = 0.01$

What is the probability that there is at least one defective integrated circuit?

Either there is at least one defective integrated circuit, that is probability $P(X > 0)$ , or there are no defective integrated circuits, that is probability $P(X = 0)$ . The sum of these probabilities is decimal 1. We want to find $P(X>0)$ .