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adelina 88 [10]
3 years ago
9

Jimmy invests $4,000 in an account that pays 2% annual interest, compounded semi-annually. What is his balance, to the nearest c

ent, at the end of 10 years?
Mathematics
1 answer:
julia-pushkina [17]3 years ago
6 0

Answer:

$4,880.80

Step-by-step explanation:

A = P(1 + r/n)^nt

Where,

A = future value

P = principal = $4,000

r = interest rate = 2% = 0.02

n = number of periods = 2

t = time = 10

A = P(1 + r/n)^nt

= 4000( 1 + 0.02/2)^2*10

= 4000(1 + 0.01)^20

= 4000( 1.01 )^20

= 4000(1.2202)

= 4,880.8

A = $4,880.80 to the nearest cents

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A science teacher has a supply of 5% hydrochloric acid and a supply of 65% hydrochloric acid(HC1). How much of each solution sho
Sphinxa [80]
Assume that the amount needed from the 5% solution is x and that the amount needed from the 65% solution is y.

We are given that, the final solution should be 42 ml, this means that:
x + y = 42 ...........> equation I
This can also be written as:
x = 42-y .......> equation II

We are also given that the final concentration should be 45%, this means that:
5% x + 65% y = 45% (x+y)
0.05x + 0.65y = 0.45(x+y)


We have x+y = 42 from equation I, therefore:
0.05x + 0.65y = 0.45(42)
0.05x + 0.65y = 18.9 .........> equation III

Substitute with equation II in equation III as follows:
0.05x + 0.65y = 18.9
0.05(42-y) + 0.65y = 18.9
2.1 - 0.05y + 0.65y = 18.9
0.6y = 18.9 - 2.1
0.6y = 16.8
y = 28 ml

Substitute with y in equation II to get x as follows:
x = 42-y
x = 42 - 28
x = 14 ml

Based on the above calculations:
amount of 5% solution = x = 14 ml
amount of 65% solution = y = 28 ml
The correct choice is:
The teacher will need 14 mL of the 5% solution and 28 mL of the 65% solution.
6 0
4 years ago
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