The time required to get a total amount of $5,900.00 with compounded interest on a principal of $5,000.00 at an interest rate of 5.75% per year 2.899 years

<h3>Compound Interest </h3>

Given Data

Principal P = $4000

Rate r= 5.75%

Final Amount A = %5900

Calculation Steps:

First, convert R as a percent to r as a decimal

r = R/100

r = 5.75/100

r = 0.0575 per year,

Then, solve the equation for t

t = ln(A/P) / n[ln(1 + r/n)]

t = ln(5,900.00/5,000.00) / ( 4 × [ln(1 + 0.0575/4)] )

t = ln(5,900.00/5,000.00) / ( 4 × [ln(1 + 0.014375)] )

t = 2.899 years

Learn more about compound interest here:

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8 0

2 years ago

A TUNCTIONS, AND SYSTEMS

SashulF [63]

Answer:

slope = -1

Step-by-step explanation:

$Y_{2} - Y_{1} / X_{2} - X_{1}$

-2 - (-6) / 3 -7

4 / -4

-1

hope this helps :)

5 0

3 years ago

Which of the following products is undefined?

mash [69]

Answer:

A is undefined for the ff answerz there

8 0

2 years ago

A sample of a radioactive substance decayed to 96.5% of its original amount after a year. (Round your answers to two decimal pla

jek_recluse [69]

Answer:

The half-life of the substance is 19.47 years.

Step-by-step explanation:

The equation for the amount of substance remaining is given by the following equation:

$Q(t) = Q(0)e^{-rt]$

In which Q(t) is the amount remaining after t years, Q(0) is the initial amount and r is the rate that this amount decreases.

A sample of a radioactive substance decayed to 96.5% of its original amount after a year.

This means that $Q(1) = 0.965Q(0)$

We use this to find r. So

$Q(t) = Q(0)e^{-rt]$

$0.965Q(0) = Q(0)e^{-r]$

$e^{-r} = 0.965$

$\ln{e^{-r}} = \ln{0.965}$

$-r = =0.0356$

$r = 0.0356$

$Q(t) = Q(0)e^{-0.0356t}$

(a) What is the half-life of the substance?

This is t when Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0356t}$

$0.5Q(0) = Q(0)e^{-0.0356t]}$

$e^{-0.0356t} = 0.5$

$\ln{e^{-0.0356t}} = \ln{0.5}$

$-0.0356t = \ln{0.5}$

$t = -\frac{\ln{0.05}}{0.0356}$

$t = 19.47$

The half-life of the substance is 19.47 years.

8 0

3 years ago

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