Colin invests £2350 into a savings account. The bank gives 4.2% compound
1 answer:
Answer:
Answer:
£3692
Step-by-step explanation:
A = p(1 + r/n)^nt
Where,
A = future value
P = principal = £2350
r = interest rate = 4.2% = 0.042
n = number of periods = 1(annual)
t = time = 4 years
A = p(1 + r/n)^nt
= 2350(1 + 0.042/1)^1*4
= 2350(1 + 0.042)^4
= 2350(1.042)^4
= 2350(1.5789)
= 2770.42
A = £2770.42
Total years = 10
Remaining years = 10 – 4
= 6 years
Remaining 6 years
P = £2770.42
r = 4.9% = 0.049
n = 1
t = 6
A = p(1 + r/n)^nt
= 2770.42(1 + 0.049/1)^1*6
= 2770.42(1 + 0.049)^6
= 2770.42(1.049)^6
= 2770.42(1.3325)
= 3691.59
A = £3691.59
Approximately £3692
Thank you!
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Step-by-step explanation:
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where

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a) 
b)
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
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
d) 