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Ugo [173]
2 years ago
5

The life of light bulbs is distributed normally. The standard deviation of the lifetime is 20 hours and the mean lifetime of a b

ulb is 500 hours. Find the probability of a bulb lasting for between 480 and 526 hours.
Mathematics
1 answer:
FinnZ [79.3K]2 years ago
5 0

Answer:

The probability of a bulb lasting for between 480 and 526 hours=0.74454

Step-by-step explanation:

We are given that

Standard deviation of the lifetime,\sigma=20hours

Mean, \mu=500hours

We have to find the probability of a bulb lasting for between 480 and 526 hours.

P(480

P(480

P(480

P(480

P(480

Hence, the probability of a bulb lasting for between 480 and 526 hours=0.74454

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Answer:

1)

Minimum is a 4th Degree

2)

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3)

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Step-by-step explanation:

Part 1)

The minimum degree of our function will be 4.

Looking at the graph, we know that the graph crosses the x-axis at -4 and -1. Since it <em>crosses through</em> the x-axis at these two points, these two factors must have an odd multiplicity.

So, it can be anything 1, 3, 5, 7, etc.

However, we will choose the lowest one, 1.

Next, we know that the graph <em>bounces off</em> at 3.

So, it must have an even multiplicity. In other words, 2, 4, 6, 8, etc.

We choose the lowest one, 2.

Therefore, the minimum degree of our function will be 1+1+2 or 4.

Part 2)

The degree of our polynomial is (and will always be) even. Therefore, both ends of the graph will go in the same direction.

Recall the simplest even polynomial, the parent quadratic function. When the leading coefficient is positive, both of the ends go straight up.

This applies to all polynomials with even degrees.

Therefore, since the arms of the graph is going straight up towards positive infinity, the leading coefficient of our graph must be positive.

Part 3)

This is similar to Part 1.

We can see that the graph touches the x-axis at -4, -3, and 1. So, the zeros of the function is: x=-4, -1, 3

We know that it<em> passes through</em> x=-4 \text{ and } x=-1 . So, these two factors must have an odd multiplicity.

However, since the graph <em>bounces off</em> x=3, this factor must have an even multiplicity.

And we're done!

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