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3 years ago

A sequence is defined by the recursive formula f (n 1) = f(n) – 2. if f(1) = 18, what is f(5)?

1 answer:

4 0

Would assume:

f(n + 1) = f(n) - 2

f(2) = f(1 + 1) = f(1) - 2 = 18 - 2 = 16

<span>f(3) = f(2 + 1) = f(2) - 2 = 16 - 2 = 14
</span>
<span>f(4) = f(3 + 1) = f(3) - 2 = 14 - 2 = 12
</span>
<span>f(5) = f(4 + 1) = f(4) - 2 = 12 - 2 = 10
</span>
f(5) = 10

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<h3>What is a range?</h3>

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<h3>How to identify the range and domain of this graph?</h3>

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Read 2 more answers

Each multiple-choice question on an exam has four choices. Suppose that there are 10 questions and the choice is selected random

Anika [276]

Answer:

It is discrete, but not uniform, since each outcome is not equally as likely.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either it is answered correctly, or it is not. The probability of answered a question is independent from the probability of answeing other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Each multiple-choice question on an exam has four choices.

This means that $p = \frac{1}{4} = 0.25$

10 questions

This means that $n = 10$

Let X denote the number of questions answered correctly. Does X have a discrete uniform distribution? Why or why not?

It is discrete, because the number of questions answered correctly is a discrete value.

However, it is not uniform. A distribution is uniform when each outcome is equally as likely. For example, the probability of answering 0 questions correctly is different of the probability of answering 1 question correctly, as we see below.

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

$P(X = 0) = C_{10,0}.(0.25)^{0}.(0.75)^{10} = 0.0563$

$P(X = 1) = C_{10,1}.(0.25)^{1}.(0.75)^{9} = 0.1877$

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