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Vadim26 [7]
2 years ago
12

What is the maximum value of this function? f(x)=−16x²+32x+20

Mathematics
1 answer:
cestrela7 [59]2 years ago
6 0

Answer: The maximum value is 35.94

Step-by-step explanation:

We have the function:

f(x) = -16*x^2 + 34*x + 20

This is a quadratic equation, and the first thing we can see is that the leading coefficient is smaller than zero, which means that the "arms" of the graph will go downwards, which means that the maximum of our function will be at the vertex.

First, we know that the vertex of a quadratic function is when f'(x) = 0.

f'(x) = 2*(-16*x) + 34 = 0.

x = -34/(-2*16) = 34/32 = 17/16.

Now we evaluate our function in the point x = 17/16.

f(17/16) = -16*(17/16)^2 + 32*(17/16) + 20 = 35.94

The maximum value is 35.94

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The water under a draw bridge was 15 feet deep at midnight. As the tide went out, the
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The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev
Vladimir [108]

Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 30.05 inches

           \sigma   = standard deviation = 0.2 inches

             n = sample of sheets = 4

So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches)  = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z < 3) = 0.99865

 P(X bar <= 30.25) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches)  = 0.99865 - 0.97725

                                                                                       = 0.0214

                                       

7 0
2 years ago
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