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2 years ago

Can someone please please help me

Mathematics

1 answer:

velikii [3]2 years ago

7 0

Answer:

order: 2, 3, 1

Step-by-step explanation:

Reasonableness checks and your knowledge of integers and fractions will help you solve this. The offered questions are intended to help you think this through.

For an output of -31, the machine (x -2)² cannot possibly be last. Its output can only be positive.

machine 2, (x-2)², cannot be last

Also, machine 3 cannot be last. For the output to be -31, the input to machine 3 must be -1/31. Neither of the other machines can produce a fraction with the inputs they might receive.

For an input of x=0, the machine 1/x cannot possibly be first. 1/0 is undefined.

The other two machines will give the following outputs for an input of 0:

machine 1: 4(0) -32 = -32

machine 2: (0 -2)² = 4

It is unlikely that machine 1 will be first, because the other two machines cannot do anything useful with -32 as an input.

machine 3, 1/x, cannot be first

The reasoning of part (a) tells you the last machine must be machine 1. The reasoning of part (b) tells you the first machine cannot be 3, so must be 2. The order of the machines must be ...

machine 2: (0 -2)² = 4 . . . . . . . using an input of 0
machine 3: 1/4 = 1/4
machine 1: 4(1/4) -32 = -31 . . . . desired output

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Using the binomial distribution, the probabilities are given as follows:

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<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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The parameters are:

x is the number of successes.

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p is the probability of a success on a single trial.

For this problem, the fixed parameter is:

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Item a:

The probability is P(X = 1) when n = 1, hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{1,1}.(0.37)^{1}.(0.63)^{0} = 0.37$

Item b:

The probability is P(X = 3) when n = 3, hence:

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$P(X = 3) = C_{3,3}.(0.37)^{3}.(0.63)^{0} = 0.5065$

Item c:

The probability is P(X = 2) when n = 4, hence:

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