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kolezko [41]
3 years ago
14

Perry Mazza wants to borrow $30,000 from the bank. The interest rate is 6% and the term is for 5 years. What is the amount of in

terest paid?
Mathematics
1 answer:
omeli [17]3 years ago
8 0

Answer: In both the case amount of interest is paid are:

1) $9000

2) $10146.76

Explanation:

Since we have given that

Principle amount = $30000

Interest rate = 6%

Number of years = 5 years

we'll use " Simple interest " first,

As we know the formula for " Simple Interest ":

I=\frac{P\times R\times T}{100}\\\\I=\frac{30000\times 6\times 5}{100}\\\\I=$9000

Now, we'll do it by "Compound Interest "

Amount=P(1+\frac{r}{100})^n\\\\Amount=30000(1+\frac{6}{100})^5\\\\Amount=$40146.76

So,

Interest = Amount-Principle\\\\Interest =40146.76-30000\\\\Interest =\$10146.76

Hence in both the case amount of interest is paid are:

1) $9000

2) $10146.76


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He number of E.coli bacteria cells in a pond of stagnant water can be represented by the function below, where A represents the
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The function that represents the number of E.coli bacteria cells per 100 mL of water as t years elapses, and is missing in the question, is:

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Answer:

   Option A. 59%

Explanation:

The base, 1.123, represents the multiplicative constant rate of change of the function, so you just must substitute 1 for t in the power part of the function::

  • A(t)=136(1.123)^{4t}

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Then, the multiplicative rate of change is 1.590, which means that every year the number of E.coli bacteria cells per 100 mL of water increases by a factor of 1.590, and that is 1.59 - 1 = 0.590 or 59% increase.

You can calculate it also using two consecutive values for t. For instance, use t =1 and t = 1

t=1\\\\ A(1)=136(1.123)^4\\\\\\t=2\\ \\ \\A(2)=136(1.123)^8\\ \\ \\A(2)/A(1)=1.123^8/1.1123^4=1.123^4=1.590

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A fire company keeps two rescue vehicles. Because of the demand on the vehicles and the chance of mechanical failure, the probab
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Answer:

(a) P (Both vehicles are available at a given time) = 0.81

(b) P (Neither vehicles are available at a given time) = 0.01

(c) P (At least one vehicle is available at a given time) = 0.99

Step-by-step explanation:

Let A = Vehicle 1 is available when needed and B = Vehicle 2 is available when needed.

<u>Given</u>:

The availability of one vehicle is independent of the availability of the other, i.e. P (A ∩ B) = P (A) × P (B)

P (A) = P (B) = 0.90

(a)

Compute the probability that both vehicles are available at a given time as follows:

P (Both vehicles are available) = P (Vehicle 1 is available) ×

                                                              P (Vehicle 2 is available)

                                  P(A\cap B)=P(A)\times P(B)

                                                  =0.90\times0.90\\=0.81

Thus, the probability that both vehicles are available at a given time is 0.81.

(b)

Compute the probability that neither vehicles are available at a given time as follows:

P (Neither vehicles are available) = [1 - P (Vehicle 1 is available)] ×

                                                                   [1 - P (Vehicle 2 is available)]

                                    P(A^{c}\cap B^{c})=[1-P(A)]\times [1-P(B)]\\

                                                       =(1-0.90)\times (1-0.90)\\=0.10\times0.10\\=0.01

Thus, the probability that neither vehicles are available at a given time is 0.01.

(c)

Compute the probability that at least one vehicle is available at a given time as follows:

P (At least one vehicle is available) = 1 - P (None of the vehicles are available)

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Thus, the probability that at least one vehicle is available at a given time is 0.99.

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