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Aleks04 [339]
3 years ago
9

Please answer asap!

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
8 0

ANSWER

The explicit rule is:

a_n=2{( \frac{1}{3} )}^{n - 1}

EXPLANATION

The recursive rule is

a_1=2

and

a_n= \frac{1}{3} a_ {n - 1}

We can rewrite to get,

\frac{a_n}{a_ {n - 1}} = \frac{1}{3}

This implies that, the constant ratio is:

r =  \frac{1}{3}

The explicit rule is given by:

a_n=a_1 {r}^{n - 1}

We substitute the values to obtain, the explicit rule as:

a_n=2{( \frac{1}{3} )}^{n - 1}

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Step-by-step explanation:

It is provided that,

<em>V</em> = a student has a Visa card

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The probability of a student having master card is:

P(M) = \frac{n(M)}{N}= \frac{32}{100}=0.32

The probability of a student having  visa card and a master card is:

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The conditional probability of an event, say A, given that another event, say B, has already occurred is,

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(a)

Compute the probability that a student has a master card given that he/she has a visa card also, i.e. P (M | V) as follows:

P(M|V)=\frac{P(V\cap M)}{P(V)} =\frac{0.12}{0.40}=0.30

Thus, the value of P (M | V) is 0.30.

(b)

Compute the probability that a student does not have a master card given that he/she has a visa card also, i.e. P (M^{c} | V) as follows:

P (M^{c} | V)=1-P(M|V)=1-0.30=0.70

Thus, the value of P (M^{c} | V) is 0.70.

(c)

Compute the probability that a student has a visa card given that he/she has a master card also, i.e. P (V | M) as follows:

P(V|M)=\frac{P(V\cap M)}{P(M)} =\frac{0.12}{0.32}=0.375

Thus, the value of P (V | M) is 0.375.

(d)

Compute the probability that a student does not have a visa card given that he/she has a master card also, i.e. P(V^{c}|M) as follows:

P(V^{c}|M)=1-P(V|M)=1-0.375=0.625

Thus, the value of P(V^{c}|M) is 0.625.

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