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murzikaleks [220]
3 years ago
8

Which expression is equivalent to (x^2 -8) - (-2x^2+4)?​

Mathematics
1 answer:
Phoenix [80]3 years ago
3 0

Answer:

D) 3x^2 - 12

Step-by-step explanation:

Using PEMDAS;

There is no need to evaluate the part of the equation (x^2 - 8) because is no need to, as it is already in its simplest form.

We must evaluate the part of the equation continuing with, "- (-2x^2+4)," as it is not in its simplest form.

Evaluating "- (-2x^2+4)":

Step 1: Distributing the negative

Once distributing the negative symbol amongst the values within the parenthesis according to PEMDAS, we get "2x^2 - 4" as the product.

Step 2: Consider the rest of the equation to evaluate

Since the part of the equation is still in play here as it is a part of the original equation to be solved, we must evaluate it as a whole to get the final answer.

Thus,

x^2 -8 + 2x^2 - 4 = ___

*we can remove the parenthesis as it has no purpose, since it makes no difference.

Evaluating for the answer, we get,

x^2+2x^2 + (-8 - 4) = 3x^2 - 12

Hence, the answer is D) 3x^2 - 12.

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2 years ago
Consider a series system composed of 4 separate components where each component has a 30% chance of failing. Assume each compone
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Answer:

16.15% probability that exactly 3 of them would function

Step-by-step explanation:

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.

Probability of each system working:

4 components, which means that n = 4

Each has a 30% probability of failing, so p = 1 - 0.3 = 0.7

For the system to work, all 4 components have to work. This is P(X = 4).

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

P(X = 4) = C_{4,4}.(0.7)^{4}.(0.3)^{0} = 0.2401

0.2401 probability of a system working.

If you have 7 of these systems, what is the probability that exactly 3 of them would function?

Now 7 systems, so n = 7

0.2401 probability of a system working.

We have to find P(X = 3).

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

P(X = 3) = C_{7,3}.(0.2401)^{3}.(0.7599)^{4} = 0.1615

16.15% probability that exactly 3 of them would function

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