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Bas_tet [7]
3 years ago
15

445 randomly selected light bulbs were tested in a laboratory, 316 lasted more than 500 30) hours. find a point estimate of the

proportion of all light bulbs that last more than 500 hours.
Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
5 0
Total number of bulbs tested, N = 445
Number of bulbs that lasted more than 500 hrs, x = 316

Point estimate of bulbs that lasted more than 500 hrs = x/N = 316/445 ≈ 0.71 = 0.71*100 = 71%
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Answer:

The work is in the explanation.

Step-by-step explanation:

The sine addition identity is:

\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).

The sine difference identity is:

\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a).

The cosine addition identity is:

\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b).

The cosine difference identity is:

\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b).

We need to find a way to put some or all of these together to get:

\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}.

So I do notice on the right hand side the \sin(a+b) and the \sin(a-b).

Let's start there then.

There is a plus sign in between them so let's add those together:

\sin(a+b)+\sin(a-b)

=[\sin(a+b)]+[\sin(a-b)]

=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]

There are two pairs of like terms. I will gather them together so you can see it more clearly:

=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]

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So this implies:

\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)

Divide both sides by 2:

\frac{\sin(a+b)+\sin(a-b)}{2}=\sin(a)\cos(b)

By the symmetric property we can write:

\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}

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2 years ago
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Answer:

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Open bracket

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Dividing by 2

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3 0
3 years ago
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