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stellarik [79]
3 years ago
6

Karen Gaines invested $12,000 in a money market account with an interest rate of 2.25% compounded semiannually. Six years later,

Karen withdrew the full amount to put toward the down payment
on a new house. How much did Karen withdraw from the account?

(Round to the nearest cent as needed.)
Mathematics
2 answers:
deff fn [24]3 years ago
8 0

Answer:

$13,724

Step-by-step explanation:

From the above question, we are given the following values:

Principal = P = $12,000

Interest = r = 2.25% = 0.0225

Compounded semi annually = n = 2

Time (t) = 6 years

From the question, we understand that we are to find the total (full) Amount Karen withdrew.

The formula to use to calculate the Total Amount of money given that this is a compound interest question is:

Total(full) Amount = P( 1 + r/n) ^n/t

= $12,000( 1 + 0.0225/2) ^2×6

= $12,000(1.01125)^12

= $13724.093289

Approximately to the nearest cent

≈ $13,724

Therefore, Karen withdrew $13,724 from the account.

Alex3 years ago
4 0

Karen Gaines invested $14000 in a money market account with an interest rate of 2.75% compounded semiannually. Six years later, Karen withdrew the full amount to put toward the down payment

Round to the nearest cent as needed.)

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Answer: Option A) 729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12} is the correct expansion.

Explanation:

on applying binomial theorem,  (a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r

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⇒ (3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6

⇒(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}

⇒(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}

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