{1800×(-90)}÷495÷(-22)

Mathematics

1 answer:

Pani-rosa [81]3 years ago

8 0

Answer:

7200

Step-by-step explanation:

{1800× (90)} ÷ 495 ÷ -(-22)

= 162000÷ 495÷ 22

= 162000×22÷ 495

= 7200

You might be interested in

Which of these is an even function?A. f(x)=5x^2-x b. f(x)=3x^3+ x c. f(x)=6x^2-8 d. f(x)= 4x^3 +2x^2

Llana [10]

I think it’s A from what I can remember

4 0

3 years ago

A playing card is chosen at random from a standard deck of cards. What is the probability of choosing the following:

lara [203]

1/52- 5 of Dimonds and 1/13- Jack

4 0

4 years ago

Your teacher announced a 50 point math quiz

77julia77 [94]

Answer:

There would be 5, 4 point questions and 10, 3 point answers

Step-by-step explanation:

4 0

3 years ago

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

neonofarm [45]

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 $P(X>0)$ .