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3 years ago

Do you know number 11 to 13

Mathematics

1 answer:

umka21 [38]3 years ago

3 0

Answer:

see explanation

Step-by-step explanation:

(11)

Note the differences in the values of the sequence

5 - 2 = 3

8 - 5 = 3

11 - 8 = 3

Since the differences are common then these are the terms of an arithmetic sequence with explicit formula

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference, thus

$a_{n}$ = 2 + 3(n - 1) = 2 + 3n - 3 = 3n - 1 ← explicit formula

(12)

A recursive formula allows the next term in the sequence to be found from the previous term.

To obtain the next term add 3 to the previous term, thus

$a_{n+1}$ = $a_{n}$ + 3 , with a₁ = 2

(13)

Using the explicit formula, then

$a_{20}$ = (3 × 20) - 1 = 60 - 1 = 59

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Answer:

a) There is n 81.92% probability that she gets an A.

b) If she gets the first problem correct, there is an 89.6% probability that she gets an A.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the answer is correct, or it is not. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

For this problem, we have that:

The probability she gets any problem correct is 0.8, so $\pi = 0.8$ .

(a) What is the probability she gets an A?

There are four problems, so $n = 4$

Jane must get at least three of the four problems on the exam correct to get an A.

So, we need to find $P(X \geq 3)$

$P(X \geq 3) = P(X = 3) + P(X = 4)$

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

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$P(X = 4) = C_{4,4}.(0.80)^{4}.(0.2)^{0} = 0.4096$

$P(X \geq 3) = P(X = 3) + P(X = 4) = 2*0.4096 = 0.8192$

There is n 81.92% probability that she gets an A.

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Now, there are only 3 problems left, so $n = 3$

To get an A, she must get at least 2 of them right, since one(the first one) she has already got it correct.

So, we need to find $P(X \geq 2)$

$P(X \geq 3) = P(X = 2) + P(X = 3)$

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$P(X \geq 3) = P(X = 2) + P(X = 3) = 0.384 + 0.512 = 0.896$

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Shkiper50 [21]

Answer:

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