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valentina_108 [34]

3 years ago

13

NEED HELP 20 points

2 answers:

garik1379 [7]3 years ago

6 0

Answer:

no solution

Step-by-step explanation:

kipiarov [429]3 years ago

4 0

The answer is D. (No solution)

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Compound interest In Exercise,$3000 is invested in an account at interest rate r,compounded continuously.Find the time required

xenn [34]

Answer:

(a) 8.15

(b) 12.92

Step-by-step explanation:

Given: P = $3000, r = 0.085

$A = Pe^{rt}$

Where

A is the Amount

P is the Principal

r is the rate

t is the time

(a) For the amount to double, A = 2 × P

A = 2 × $3000

A = $6000

$6000 = 3000e^{0.085t}$

$\frac{6000}{3000} = e^{0.085t}$

$2 = e^{0.085t}$

Take $log_{e}$ of both sides

$log_{e}2 = log_{e}e^{0.085t}$

But $log_{e}e = 1$

∴ $ln2 = 0.085t$

$t = \frac{ln2}{0.085}$

$t = \frac{0.693}{0.085}$

t = 8.15

(b) For the amount to double, A = 3 × P

A = 3 × $3000

A = $9000

$9000 = 3000e^{0.085t}$

$\frac{9000}{3000} = e^{0.085t}$

$3 = e^{0.085t}$

Take $log_{e}$ of both sides

$log_{e}3 = log_{e}e^{0.085t}$

But $log_{e}e = 1$

∴ $ln3 = 0.085t$

$t = \frac{ln3}{0.085}$

$t = \frac{1.0986}{0.085}$

t = 12.92

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3 years ago