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3 years ago

What is the 20th term of the arithmetic sequence?

Mathematics

2 answers:

podryga [215]3 years ago

8 0

You just need to substitute 20 for x and then solve.

a(20) = -5 +(20-1)3

Simplify: a(20) = -5 + (60-3)

Simplify: a(20) = -5 + 57

Result: a(20) = 52

So, 52 is the 20th term of the arithmetic sequence.

Mandarinka [93]3 years ago

8 0

In the equation
$a(n) = - 5 + (n - 1)3$
Let n=20 and evaluate it as:

$a(20) = - 5 + (20 - 1)3 \\ \\ a(20) = - 5 + (19)3 \\ \\ a(20) = - 5 + 57 \\ \\ a(20) = 52$

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A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a

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Answer:

$P(X \geq 5)=1-P(X$

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=19, p=0.3)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

And we want to find this probability:

$P(X \geq 5)$

And we can use the complement rule:

$P(X \geq 5)=1-P(X$

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

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