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mr Goodwill [35]

2 years ago

11

Sylvia invested $500 in an account compounded annually with an interest rate of 8%. Manuel invested $600 in an account with a co

mpound interest rate of 7.25%. Using the rule of 72, mc012-1.jpg, who will double their money first?
Sylvia will double her money first, in approximately 9 years.
Manuel will double his money first, in approximately 10 years.
Manuel will double his money first, in approximately 9 years.
Sylvia will double her money first, in approximately 10 years.

2 answers:

Sav [38]2 years ago

8 0

Syliva
T=72÷8=9 years

Manul
T=72÷7.25=9.9=10 years

Option A

Anit [1.1K]2 years ago

7 0

Answer: Sylvia will double her money first, in approximately 9 years.

Step-by-step explanation:

By the rule of 72,

An amount is doubled with a rate of interest in a certain time when the product of the annual rate and time(in years) is equal to 72,

In the Sylvia's investment,

Rate of interest = 8 %

Let the time in which the investment doubled = x years,

Thus, by the above rule,

8 x = 72

⇒ x = 9 years,

In the Manuel's investment,

The rate of interest = 7.25

Let this investment is doubled in y years,

Thus, by the above rule,

7.25 y = 72

⇒ y = 9.93103448 ≈9.93 years

Hence, Sylvia's investment takes less time to be doubled,

⇒ Option A is correct.

Note: The rule of 72 does not always give the exact answer,

That's why number of years found by rule of 72 for each case will be said approximately.

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a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

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Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. 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.

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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.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

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$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

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$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

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d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

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