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Natasha2012 [34]
3 years ago
5

Use the following explicit function to identify f(1) & r, and list the first five terms of the sequence.

Mathematics
2 answers:
shusha [124]3 years ago
7 0

Given:

The explicit function is:

To find:

and first 5 terms.

Solution:

The explicit formula a geometric sequence is:

           ...(i)

Where, a is the first term and r is the common ratio.

We have,

        ...(ii)

On comparing (i) and (ii), we get

And,

For ,

For ,

Similarly, substituting , we get

Therefore,  and the first five terms are .

rjkz [21]3 years ago
5 0

Given:

The explicit function is:

f(n)=1000\left(\dfrac{1}{4}\right)^{n-1}

To find:

f(1)=?,r=? and first 5 terms.

Solution:

The explicit formula a geometric sequence is:

f(n)=ar^{n-1}            ...(i)

Where, a is the first term and r is the common ratio.

We have,

f(n)=1000\left(\dfrac{1}{4}\right)^{n-1}         ...(ii)

On comparing (i) and (ii), we get

a=1000

f(1)=1000

And,

r=\dfrac{1}{4}

For n=1,

f(1)=1000\left(\dfrac{1}{4}\right)^{1-1}

f(1)=1000\left(\dfrac{1}{4}\right)^{0}

f(1)=1000(1)

f(1)=1000

For n=2,

f(2)=1000\left(\dfrac{1}{4}\right)^{2-1}

f(2)=1000\left(\dfrac{1}{4}\right)^{1}

f(2)=1000\times \dfrac{1}{4}

f(2)=250

Similarly, substituting n=3,4,5, we get

f(3)=62.5

f(4)=15.625

f(5)=3.90625

Therefore, f(1)=1000,r=\dfrac{1}{4} and the first five terms are 1000,250,62.5,15.625,3.90625.

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Assume that a procedure yields a binomial distribution with a trial repeated n = 5 times. Use some form of technology to find th
Digiron [165]

Answer:

P(X = 0) = 0.0263

P(X = 1) = 0.1407

P(X = 2) = 0.3012

P(X = 3) = 0.3224

P(X = 4) = 0.1725

P(X = 5) = 0.0369

Step-by-step explanation:

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.

In this problem we have that:

n = 5, p = 0.517

Distribution

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

P(X = 0) = C_{5,0}.(0.517)^{0}.(0.483)^{5} = 0.0263

P(X = 1) = C_{5,1}.(0.517)^{1}.(0.483)^{4} = 0.1407

P(X = 2) = C_{5,2}.(0.517)^{2}.(0.483)^{3} = 0.3012

P(X = 3) = C_{5,3}.(0.517)^{3}.(0.483)^{2} = 0.3224

P(X = 4) = C_{5,4}.(0.517)^{4}.(0.483)^{1} = 0.1725

P(X = 5) = C_{5,5}.(0.517)^{5}.(0.483)^{0} = 0.0369

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