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Paladinen [302]
3 years ago
6

A shipment of 8 computers contains 3 with defects. Find the probability that a sample of size 2, drawn from the 8, will not cont

ain a defective computer.
What is the probability that a sample of 2 of the 8 computers will not contain a defective computer?
(Type an integer or a simplified fraction.)
Mathematics
1 answer:
liq [111]3 years ago
6 0

Answer:

5/28

Step-by-step explanation:

First of all, probability is defined as the number of ways a certain event can happen divided by the total ways of an event happening. In this scenario, we are asked to find the number of non-defective computers divided by the total ways that a sample of two can be chosen.

The number of ways to choose 2 computers from 8 can be written as 8\choose2, which is equal to \frac{8\cdot7}{2}, which is 28. Now, there are 3 defective computers, for a total of 5 non-defective computers, so the probability is 5/28.

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drawbridge at the entrance to an ancient castle is raised and lowered by a pair of chains. The figure represents the drawbridge
jasenka [17]

Answer:

4.0 meters, ∠C = 39°, ∠A = 51°

Step-by-step explanation:

Firstly, our diagram shows us that the given triangle is actually a right triangle. So we can use the <em>Pythagorean Theorem</em> to solve for the height of the chain:

a^{2} +b^{2} =c^{2}

(3.3)^{2} +b^{2} =(5.2)^{2}

b^{2} =(5.2)^{2}-(3.3)^{2}

b =\sqrt{(5.2)^{2}-(3.3)^{2}}

b=4.0187...

b=4.0 m

Now, we can use the <em>Law of Cosines</em> to figure out one of the acute angles:

c^{2}  =a^{2} +b^{2} -2ab(cos(C))

(3.3)^{2}  =(4.0)^{2} +(5.2)^{2} -2(4.0)(5.2)(cos(C))

cos(C)=\frac{(3.3)^{2}-(4.0)^{2} -(5.2)^{2}}{-2(4.0)(5.2)}

C=cos^{-1}( \frac{(3.3)^{2}-(4.0)^{2} -(5.2)^{2}}{-2(4.0)(5.2)})

C=39.3915...

∠C = 39°

And since we know that all angles in a triangle add up to 180°:

∠A + ∠B + ∠C = 180

∠A + 90 + 39 = 180

∠A = 180 - 90 - 39

∠A = 51°

However, you should always review any answers on the Internet and make sure they are correct! Check my work to see if I made any mistakes!

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