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2 years ago

Michael invests USD 20 000 at 9.6% p.a. compounded monthly.

Mathematics

1 answer:

alexandr1967 [171]2 years ago

5 0

Given :

A = 25000

P = 20000

r % = 9.6 % = 0.096

n = 12

To Find :

The time taken say t.

Solution :

We know, compound interest is given by :

$A=P(1+\dfrac{r}{n})^{n.t}$

Taking log both sides :

$A=P(1+\dfrac{r}{n})^{n.t}\\\\log\ \dfrac{A}{P}= n.t\times log( 1+\dfrac{r}{n})\\\\t =\dfrac{1}{n}\times \dfrac{log\ \dfrac{A}{P}}{log(1+\dfrac{r}{n})}\\\\\\t=\dfrac{1}{12}\times \dfrac{log\ \dfrac{25000}{20000}}{log(1+\dfrac{0.096}{12})}\\\\\\t=2.33\ years$

Hence, this is the required solution.

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Which recursive formula can be used to generate the sequence below, where f(1) = 6 and n ≥ 1?

Trava [24]

f (n + 1) = f(n) – 5 is the recursive formula can be used to generate the sequence below, where f(1) = 6 and n ≥ 1

<h3>Solution:</h3>

Given that,

f(1) = 6 and n ≥ 1

Given sequence is 6, 1, -4, -9, -14

Let us first analyse the logic used in this sequence

6 - 5 = 1

1 - 5 = -4

-4 - 5 = -9

-9 - 5 = -14

Thus the next terms in sequence are obtained by subtracting 5 from previous term

Thus a recursive formula can be formed as:

f (n + 1) = f(n) – 5

Where "n" is the nth term

Let us check our recursive formula:

f(1+ 1) = f(1) - 5

f(2) = f(1) - 5

f(2) = 6 - 5 = 1

Thus we have got f(2) = 1 which is correct as per given sequence

4 0

2 years ago

Read 2 more answers

Find an upper limit for the zeroes 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0

erik [133]

Answer-

2 is the upper limit for the zeros.

Solution-

The given function f(x) is,

$2x^4 -7x^3 + 4x^2 + 7x - 6 = 0$

For calculating the zeros,

$\Rightarrow f(x)=0$

$\Rightarrow 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0$

$\Rightarrow 2x^4-4x^3-3x^3+6x^2-2x^2+ 4x+3x-6=0$

$\Rightarrow 2x^3(x-2)-3x^2(x-2)-2x(x-2)+3(x-2)=0$

$\Rightarrow (x-2)(2x^3-3x^2-2x+3)=0$

$\Rightarrow (x-2)(x^2(2x-3)-1(2x-3))=0$

$\Rightarrow (x-2)(x^2-1)(2x-3)=0$

$\Rightarrow (x-2)(x+1)(x-1)(2x-3)=0$

$\Rightarrow x-2=0,\ x+1=0,\ x-1=0,\ 2x-3=0$

$\Rightarrow x=2,\ x=-1,\ x=1,\ x=\frac{3}{2}$

From all the 4 roots, it can be obtained that 2 is the greatest zero.

7 0

1 year ago

Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang

kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

$P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164$

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

$P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145$

5 0

2 years ago

A student is raising money for cancer research. A local business agrees to donate an additional 25% of what the student raises,

bixtya [17]

Answer: 585 maybe??

Step-by-step explanation:

450/100 = 4.5

4,5 = 1 percent of the total amount.

4.5 x 25 = 112.5

112 = 25 percent of 450

4.5 x 5 = 22.5

22.5 = 5 percent of 450

Hope u understand and that this helps:)

4 0

1 year ago

susan took two tests.the probability of her passing both tests is 0.6.the probability of her passing the first test is 0.8.what

OverLord2011 [107]

Formula for this is as follows:
probability of her passing both 0.6/0.8 - first test and this is a fraction. 0.6/0.8
0.6/0.8= divide 0.6 by 0.8=0.75
that means probability of her passing the second test is 75%

6 0

1 year ago