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3241004551 [841]
3 years ago
11

You deposit $7000 into a bank account that has 6% interest compounded SEMIANNUALLY. Determine the amount of money in the bank af

ter 9 years.
Mathematics
1 answer:
Vlada [557]3 years ago
4 0

Answer:

$11917.03

Step-by-step explanation:

Use Compound Interest Formula:

A = 7,000 (1 + 0.06/2)^2(9)

Simplify:

A = 7,000(1.03)^18

Solve:

A = $11917.03

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A very old vending machine accepts only nickels (n) and dimes (d). Candy costs up to $0.50, but sometimes the machine will dispe
matrenka [14]
Let n be the number of nickels.
Let d be the number of dimes.
Our inequality is
10d + 5n ≤ 50

Have an awesome day! :)
8 0
3 years ago
Arianna is saving money for college so she invests $1,000 into a savings account that earns interest every year. If the amount i
Verdich [7]

Answer: $1,624

Step-by-step explanation:

The amount after x-years is given as :

9x^{2} + 50x + 1000

Since x represents number of years , this means that after 6 years , the amount will be :

9(6^{2} ) + 50(6) + 1000

⇒9(36) + 300 + 1000

⇒324 + 300 + 1000 =

Therefore, the amount after 6 years will be $1,624

7 0
3 years ago
Can someone help me pleaseeeeeee :( no links pleaseee
Sergeu [11.5K]

a) Keith's car will have less gas.

Locate the point in which miles driven = 90 (Between 60 and 120)

Since Keith's line is lower than Carmen, he has less gas.

We can also find the difference: 11 - 10 = 1.

It will have 1 less gallon.

b) Look at the point in which both lines meet. This means that they have the same amount of gallons remaining (8) at 180 miles driven.

Before 180 miles, Keith's car will always have less car remaining.

3 0
3 years ago
Find the least common denominator (LCD) of 11/5 and 3/10
lora16 [44]

Step-by-step explanation:

xtxjsogkgxhckhcfjxgxkcgkkcgkhckxglx

5 0
3 years ago
Read 2 more answers
Two pipes are used to fill a 300-gallon tub with water. The wide pipe can fill the tub in x hours. The narrow pipe can fill the
Hoochie [10]

Answer:

\frac{6}{x}-\frac{7}{y}=\frac{7}{12}

Step-by-step explanation:

#We know that the filling taps will run for a cumulative 4hrs each while the draining hole will empty for a combined 7hrs.

Let x be the rate of the first filling tap, and 2x of the 2nd filling tap and y be the rate of the hole.

#The rate at which the tub fills per hour is :

V_{tub}=\frac{1}{x}+\frac{1}{2x}-\frac{1}{y}\\\\=\frac{1}{x}(1+\frac{1}{2})-\frac{1}{y}\\\\=\frac{3}{2x}-\frac{1}{y}

#The volume of the tub at the instant moment the leak is repaired is expressed as:

V_{tub}=\frac{3}{2x}t_1-\frac{1}{y}t_2=\frac{7}{12}, \ \ \ \ \ \ t_1=4,t_2=7\\\\\\\frac{3}{2x}\times 4-\frac{1}{y}\times7\frac{7}{12}\\\\\\\frac{6}{x}-\frac{7}{y}=\frac{7}{12}

Hence, the volume of the tub after the leak is repaired is expressed as \frac{6}{x}-\frac{7}{y}=\frac{7}{12}

8 0
3 years ago
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