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3 years ago

Find the domain. p(x) = x^2+ 2

Mathematics

1 answer:

MrRissso [65]3 years ago

4 0

Answer:

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

( − ∞ , ∞ )

Set-Builder Notation:

{ x | x ∈ R }

Step-by-step explanation:

hope that helps bigger terms

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Solve this equationn 5w=17.5

Alchen [17]

Answer:

w = 3.5

Step-by-step explanation:

w = 3.5?

4 0

3 years ago

Given P(A and B) 0.20, P(A) 0.49, and P(B) = 0.41 are events A and B independent or dependent? 1) Dependent 2) Independent

blsea [12.9K]

Answer: The correct option is (1) Dependent.

Step-by-step explanation: For two events, we are given the following values of the probabilities :

P(A ∩ B) = 0.20, P(A) = 0.49 and P(B) = 0.41.

We are to check whether the events A and B are independent or dependent.

We know that

the two events C and D are said to be independent if the probabilities of their intersection is equal to the product of their probabilities.

That is, P(C ∩ D) = P(C) × P(D).

For the given two events A and B, we have

$P(A)\times P(B)=0.49\times0.41=0.2009\neq P(A\cap B)=0.20\\\\\Rightarrow P(A\cap B)\neq P(A)\times P(B).$

Therefore, the probabilities of the intersection of two events A and B is NOT equal to the product of the probabilities of the two events.

Thus, the events A and B are NOT independent. They are dependent events.

Option (1) is CORRECT.

8 0

4 years ago

) A bin contains 60 light bulbs of which 5 are defective. If 4 light bulbs are randomly selected from the bin with replacement,

8090 [49]

Answer:

0.706

Step-by-step explanation:

In this question, we are asked to calculate the probability that if 4 light bulbs are selected out of a total of 60, with 5 defective ones, all are good.

Firstly we need to find the probability of selecting a bulb that is not defective. Let’s call this P(g).

Since 5 are defective, it means 60-5 = 55 bulbs are not defective.

The probability of selecting a bulb not defective would be 55/60 = 11/12

Now, four are selected with claims that non are defective, the probability here would thus be 11/12 * 11/12 * 11/12 * 11/12 = 0.706066743827

To the nearest thousandth, we have 0.706

3 0

3 years ago

Covert 77 1/2% to a fraction in lowest term

Elodia [21]

$77\frac{1}{2} \%$ converted to fraction in lowest terms is $\frac{31}{40}$