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3 years ago

A sum of money is invested at 12% compounded quarterly. About how long will it take for the amount of money to double?

2 answers:

4 0

I'll use my own formula (similar to yours tho)

$A=P(1+ \frac{r}{n})^{nt}$
A=future amount
p=invested
r=rate in decimal
n=number of times per year compounded
t=time in years

so
when will A=2P
subsitute 2P for A
also quarterly means 4 times a year or n=4
and 12%=0.12=r
so

$2P=P(1+ \frac{0.12}{4})^{(4*t)}$
divide both sides by P
$2=(1+ \frac{0.12}{4})^{(4*t)}$
simplify
$2=(1+ 0.03)^{4t}$
$2=(1.03)^{4t}$
take the ln of both sides
$ln(2)=ln(1.03)^{4t}$
$ln(2)=4tln(1.03)$
$ln(2)=t(4ln(1.03))$
divide both sides by 4ln(1.03)
$\frac{ln(2)}{4ln(1.03)}=t$
use calculator
5.86244=t
about 5.9 years

Dmitriy789 [7]3 years ago

4 0

Answer:

5.9

Step-by-step explanation:

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2 years ago

find the equation of a line that passes through the point (4, 3) and is parallel to the line 2x-2y=11

Sonja [21]

Answer:

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Step-by-step explanation:

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1 year ago

The following is a linear programming formulation of a labor planning problem. There are four overlapping shifts, and management

dexar [7]

Answer:

d. 15

Step-by-step explanation:

Putting the values in the shift 2 function

X1 + X2 ≥ 15

where x1= 13, and x2=2

13+12≥ 15

15≥ 15

At least 15 workers must be assigned to the shift 2.

The LP model questions require that the constraints are satisfied.

The constraint for the shift 2 is that the number of workers must be equal or greater than 15

This can be solved using other constraint functions e.g

Putting X4= 0 in

X1 + X4 ≥ 12

gives

X1 ≥ 12

Now Putting the value X1 ≥ 12 in shift 2 constraint

X1 + X2 ≥ 15

12+ 2≥ 15

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this does not satisfy the condition so this is wrong.

Now from

X2 + X3 ≥ 16

Putting X3= 14

X2 + 14 ≥ 16

gives

X2 ≥ 2

Putting these in the shift 2

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Which gives the same result as above.

6 0

3 years ago

Fourth degree and a single zero of -3

Darya [45]

The degree of a polynomial is the highest power of the polynomial.

The polynomial is $\mathbf{P(x) = (x + 3)^4}$

The degree is given as:

$\mathbf{n = 4}$

The zero is given as:

$\mathbf{x = -3}$

Add 3 to both sides

$\mathbf{x + 3= 0}$

From the question, we understand that the polynomial has a single zero.

So, the polynomial is:

$\mathbf{P(x) = (x + 3)^n}$

Substitute 4 for n

$\mathbf{P(x) = (x + 3)^4}$

Hence, the polynomial is $\mathbf{P(x) = (x + 3)^4}$