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nika2105 [10]
3 years ago
15

Eric wants to get an output of 0. Can he do this with each machine? If so, how? If not, why not?

Mathematics
1 answer:
natita [175]3 years ago
4 0

Answer:

  • <u>No, he can get an output of 0 with the second machine (function B) but he cannot get an output of 0 with the first machine (function A).</u>

Explanation

The way each machine works is given by the expression (function) inside it.

<u>1)  </u><em><u>Function A</u></em>

To get an output of 0 with the function y = x² + 3, you must solve the equation x² + 3 = 0.

Since x² is zero or positive for any real number, x² + 3 will never be less than 3 (the minimum value of x² + 3 is 3). So, it is not possible to get an output of 0 with the first machine.

<u>2) </u><em><u>Function B</u></em>

Solve y=\sqrt{x}-2=0

  • \sqrt{x}-2=0

  • \sqrt{x} =2

  • x=2^{2}

  • x=4

So, he can get an output of 0 by using x = 4.

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The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

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Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have  a seat, we have  

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 =  5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

 z = \frac{x-np}{\sqrt{np(1-p)}  } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} } =  - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = \sqrt{npq} = 2.233

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