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2 years ago

How do you solve this math problem

Mathematics

2 answers:

LiRa [457]2 years ago

6 0

First, we need to simplify your quadratic equation, in the form of ax^2-bx-c=0. Your equation turns out to be x^2-4x+5=0
To find your solutions, x1+x2=-b/a or 4 in this case, and x1*x2=c/a or 5 in this case. So we need to find two numbers when added together is 4 and multiplied together to get 5. We can easily find our roots by using the roots formula, or (-b (+/-)sqrt(b^2-4ac))/2a. When plugging everything in, we have (4(+/-)sqrt(16-20)/2. As you can make out from the plugged in equation, you can't have a negative number under the radical symbol in order for it to be true.
Your answer is C, no real solutions.
Hope this helps.

zhuklara [117]2 years ago

3 0

The expressions is undefined in the set of the real numbers.

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1 year ago

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kifflom [539]

Answer:

The expected number of tests, E(X) = 6.00

Step-by-step explanation:

Let us denote the number of tests required by X.

In the case of 5 individuals, the possible value of x are 1, if no one has the disease, and 6, if at least one person has the disease.

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Case 2: P(X=6) = 1- P(X=1)

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2 years ago

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2 years ago

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\\
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\\
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