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s2008m [1.1K]
4 years ago
9

A flashlight battery manufacturer makes a model of battery whose mean shelf life is three years and four months, with a standard

deviation of three months. The distribution is approximately normal. One production run of batteries in the factory was 25,000 batteries. How many of those batteries can be expected to last between three years and one month and three years and seven months?
Mathematics
1 answer:
olga_2 [115]4 years ago
7 0

Answer:

17000 batteries

Step-by-step explanation:

Three years and one month is equivalent to the mean minus one standard deviation.

Three years and seven months is equivalent to the mean plus one standard deviation.

For a normal distribution, we know that 68% of population is between mean ± 1 sd, then can be expected that 25000*68% = 17000 of batteries last between three years and one month and three years and seven months

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Find the slope of the line that contains the points (6, 5) and (8,7)
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The slope of the line is 1

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Read 2 more answers
If g is a differentiable function such that g(x)<0 for all real numbers x and if f'(x) = (x2 - 4)g(x), which of the following
galina1969 [7]

We want to see what can we say about f(x) extremes for the given information. We will see that the correct option is:

V: f has a relative minimum at x=-2 and a relative maximum at x=2.

<h3>Maximums and minimums.</h3>

A given function f(x) will have a maximum/minimum at x₀ if:

f'(x₀) = 0

  • It will be a maximum if f''(x₀) < 0
  • It will be a minimum if f''(x₀) > 0.

Here we do know that:

  • f'(x) = (x^2 - 4)*g(x)
  • g(x) < 0 for all numbers x.

Then the only zeros of f'(x) are when (x^2 - 4) = 0.

And this happens for x = 2 and x  = -2

Now let's see if these are minimums or maximums, we have:

f''(x) = 2*x*g(x) + (x^2 - 4)*g'(x)

If we evaluate this in x = 2, we get:

f''(2) = 2*2*g(2) + (2^2 - 4)*g'(2)

       = 4*g(2) + 0*g'(2) = 4*g(2)

And g(x) is always negative, then 4*g(2) < 0, then:

f''(2) < 0

Meaning that in x = 2 we have a relative maximum.

For x = -2 we have:

f''(-2) = 2*-2*g(-2) + ((-2)^2 - 4)*g'(-2)

       = -4*g(2) > 0

Then f''(-2) > 0, meaning that in x = -2 we have a relative minimum.

Then the correct option is:

V:  f has a relative minimum at x=-2 and a relative maximum at x=2.

If you want to learn more about maximums and minimums, you can read:

brainly.com/question/1938915

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