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dmitriy555 [2]
3 years ago
5

Suppose $12000 is invested into a savings account for 23 years. The interest on the account is conpounded continuously. How much

money in interest did the account make when thr account is $45,000?
Mathematics
1 answer:
KATRIN_1 [288]3 years ago
5 0
I'm guessing to probably divide $45,000 by 23 and then multiply that answer by $12,000. I hope this helps
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An Interesting Equation of Order n: xy" - (x + n)y' + ny = 0: u(x) = e^x. (a) Prove that the given solution is indeed a solution
Igoryamba

Solution:

The given differential equation is

x y" -(x+n)y'+ny=0------------(1)

Let, y'=t

y"=t'

\frac{dy}{dx}=t\\\\y=t x

Substituting the value of , y', y'' and y in equation (1)

→x t' -(x+n)t+n t x=0

→ x t' = x t+ n t- nt x

→ x t'=t(x+n-nx)

\rightarrow \frac{t'}{t}=\frac{x+n-nx}{x}\\\\\rightarrow \frac{dt}{t}=(1-n+\frac{n}{x}) dx\\\\ \text{Integrating both sides}}\\\\ \int{\frac{dt}{t}}=\int {(1-n+\frac{n}{x}) dx}\\\\ \log t=x - nx+n \log x+\log K\\\\ \log t -\log x^n-\log K=x(1-n)\\\\\log \frac{t}{Kx^n}=x(1-n)\\\\t=Kx^n \times e^{x(1-n)}\\\\y'=Kx^n \times e^{x(1-n)}

which is a solution of Differential equation.

(b)

\frac{dy}{dx}=Kx^n\times e^{x(1-n)}\\\\ dy=Kx^n\times e^{x(1-n)} dx

Integrating both sides

y=\frac{Kx^n\times e^{x(1-n)}}{1-n}-\frac{Kn}{1-n}\int{x^{n-1}e^{x(1-n)} dx

required linear independent solution of Differential equation.

7 0
3 years ago
Jo and David picked 92 quarts of berries at King's Orchard. If David picked 6 more quarts than Jo picked, how many quarts
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Answer:

43 quarts of berries

Step-by-step explanation:

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WILL GIVE BRAINEST <br>Leo had $91, which is 7 times as much......
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C) because the 7 and x right next to each other represents multiplication. so it's:
7x=91

7X(Allison's money)=91
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3 years ago
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We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely. (a) Given that the roll results
boyakko [2]

Answer:

a) \frac{1}{7}

b) \frac{2}{15}

Step-by-step explanation:

Given : We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely.

The outcomes are :

(1, 1) (1, 2)  (1, 3)  (1, 4)  (1, 5)  (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

(a) Given that the roll results in a sum of 3 or more, find the conditional probability that doubles (the first and the second rolls result in the same number) are rolled.

Let P(A) be the probability of event that the roll results in a sum of 3 or more.

Except (1,1) rest sum is 3 or greater than 3.

So, P(A)=\frac{35}{36}

Now, P(A and B) is that the roll results in a sum of 3 or more and that doubles (the first and the second rolls result in the same number) are rolled.

i.e. (2,2), (3,3), (4,4), (5,5), (6,6) - 5

P(A\cap B)=\frac{5}{36}

The conditional probability is given by,

P(B|A)=\frac{P(A\cap B)}{P(A)}

P(B|A)=\frac{\frac{5}{36}}{\frac{35}{36}}

P(B|A)=\frac{5}{35}

P(B|A)=\frac{1}{7}

(b) Given that the two dice land on different numbers, find the conditional probability that the sum is 5.

Let P(A) be the probability of event that two dice land on different numbers.

Except (1,1),(2,2), (3,3), (4,4), (5,5), (6,6) rest two dice land on different numbers.

So, P(A)=\frac{30}{36}

Now, P(A and B) is that two dice land on different numbers and the sum is 5.

i.e. (1,4), (2,3), (3,2), (4,1) - 4

P(A\cap B)=\frac{4}{36}

The conditional probability is given by,

P(B|A)=\frac{P(A\cap B)}{P(A)}

P(B|A)=\frac{\frac{4}{36}}{\frac{30}{36}}

P(B|A)=\frac{4}{30}

P(B|A)=\frac{2}{15}

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the side lengths of a triangle are increased by 50% to make a similar triangle. does the area increase by 50% as well?
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No. The new area is 2.25 times the original area. That's (1.5 times 1.5).
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