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3 years ago

What is the solution set for 7x – 3 = 18, given the replacement set {0, 1,2,3,4} ? O

Mathematics

1 answer:

yawa3891 [41]3 years ago

4 0

Answer:

X = 3

Step-by-step explanation:

Given the expression :7x - 3 = 18

To obtain the solution set ; we solve for X in the equation ;

7x - 3 = 18

Add 3 to both sides

7x - 3 + 3 = 18 + 3

7x = 21

Divide both sides by 7

7x / 7 = 21 / 7

x = 3

Hence, x = 3

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Step-by-step explanation:

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3 years ago

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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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3 years ago

309,099,990 expanded form

Marrrta [24]

Three hundred nine million ninety nine thousand nine hundred and ninety.

5 0

3 years ago

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Under certain conditions, the number of bacteria present in a colony is approximated by f(t) = Age 0.023t, where t is in minutes

AleksandrR [38]

Answer:

5 min: 3,029,058
10 min: 3,398,220
60 min: 10,732,234

Step-by-step explanation:

The given function is evaluated by substituting the given values of t. This requires using the exponential function of your calculator with a base of 'e'. Many calculators have that value built in, or have an e^x function (often associated with the Ln function).

<h3>5 minutes</h3>

The number of bacteria present after 5 minutes is about ...

f(5) = 2.7×10^6×e^(0.023×5) ≈ 3,029,058

<h3>10 minutes</h3>

The number of bacteria present after 10 minutes is about ...

f(10) = 2.7×10^6×e^(0.023×10) ≈ 3,398,220

<h3>60 minutes</h3>

The number of bacteria present after 60 minutes is about ...

f(60) = 2.7×10^6×e^(0.023×60) ≈ 10,732,234