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tia_tia [17]
3 years ago
7

Is 8.4 a rational number

Mathematics
2 answers:
Murljashka [212]3 years ago
6 0
Yep.

All rational numbers are terminating or repeating decimals (this one is terminating because it ends). This number can also be written as a ratio of two integers (84/10).
Valentin [98]3 years ago
4 0
I think 8.4 is a rational number
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Make x the subject of the following formula x^2+3=h <br> 2x=2y=p<br> s=x^2-3
madam [21]

Answer:

x^2+3=h

to make x the subject of the formula,move 3 to the other side making it negative

x^2 = h + 3

to take out the square,take the square root of both sides.

√x^2 = √ h + 3

x = √ h + 3.

you can ask me questions privately when confused .

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3 years ago
The probability that a lab specimen contains high levels of contamination is 0.14. A group of 4 independent samples are checked.
Shtirlitz [24]

Answer:

a) 0.5470 = 54.70% probability that none contain high levels of contamination.

b) 0.3562 = 35.62% probability that exactly one contains high levels of contamination.

c) 0.4530 = 45.30% probability that at least one contains high levels of contamination.

Step-by-step explanation:

For each sample, there are only two possible outcomes. Either they contain high levels of contamination, or they do not. The samples are independent. So we use the binomial probability distribution to solve this question.

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}.p^{x}.(1-p)^{n-x}

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

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

And p is the probability of X happening.

The probability that a lab specimen contains high levels of contamination is 0.14.

This means that p = 0.14

A group of 4 independent samples are checked.

This means that n = 4

(a) What is the probability that none contain high levels of contamination?

This is P(X = 0)

P(X = 0) = C_{4,0}.(0.14)^{0}.(0.86)^{4} = 0.5470

0.5470 = 54.70% probability that none contain high levels of contamination.

(b) What is the probability that exactly one contains high levels of contamination?

This is P(X = 1)

P(X = 1) = C_{4,1}.(0.14)^{1}.(0.86)^{3} = 0.3562

0.3562 = 35.62% probability that exactly one contains high levels of contamination.

(c) What is the probability that at least one contains high levels of contamination?

Either none of the samples contain high levels of contamination, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want to find P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.5470 = 0.4530

0.4530 = 45.30% probability that at least one contains high levels of contamination.

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