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yKpoI14uk [10]
3 years ago
13

What is the range of the function y=3 Sqrt x+8

Mathematics
2 answers:
Goshia [24]3 years ago
5 0

Answer:

range is -∞<y<∞

Step-by-step explanation:

y=\sqrt[3]{x+8}

Range of the given function is same as the domain of the inverse function

LEts find the inverse for the given equation

Swap the variables x and y . then solve for y

y=\sqrt[3]{x+8}

x=\sqrt[3]{y+8}

Take cube on both sides

x^3= y+8

Subtract 8 from both sides

y=x^3-8

Inverse function is a cubic function. For cubic function the domain is set of allr eal numbers. -∞<x<∞

Range of the given function is same as the domain of the inverse function

So range is -∞<y<∞

ValentinkaMS [17]3 years ago
4 0

Answer:

(-\infty,+\infty)

Step-by-step explanation:

The given function is

y=\sqrt[3]{x+8}

The range refers to the values of y for which  x is defined.

We solve for x to obtain;

y^3=x+8

x=y^3-8

This is a polynomial function in y.

x is defined for all values of y.

Therefore the range is all real numbers.

or

(-\infty,+\infty)

The first choice is correct.

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4 0
3 years ago
Approximately 5% of calculators coming out of the production lines have a defect. Fifty calculators are randomly selected from t
baherus [9]

Answer:

0.2611 = 26.11% probability that exactly 2 calculators are defective.

Step-by-step explanation:

For each calculator, there are only two possible outcomes. Either it is defective, or it is not. The probability of a calculator being defective is independent of any other calculator, which means that the binomial probability distribution is used 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.

5% of calculators coming out of the production lines have a defect.

This means that p = 0.05

Fifty calculators are randomly selected from the production line and tested for defects.

This means that n = 50

What is the probability that exactly 2 calculators are defective?

This is P(X = 2). So

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

P(X = 2) = C_{50,2}.(0.05)^{2}.(0.95)^{48} = 0.2611

0.2611 = 26.11% probability that exactly 2 calculators are defective.

3 0
3 years ago
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