We use the formula:
Total = Principal * e ^ (rate * years)
Total = 1,500 * e ^ (.03 * 15)
Total = 1,500 * e ^ .45
Total = 1,500 *
<span>
<span>
<span>
1.5683121854
</span>
</span>
</span>
Total =
<span>
<span>
<span>
2,352.47
</span>
</span>
</span>
Answer:
2.2%
Step-by-step explanation:
Given the following :
Population in year 2000 (A) = 4.2 million
Expected population every 32 years = 2 *A
The growth rate per year =?
The population figure after 32 years = (2 * 4.2 million) = 8.4 million
Using the exponential growth formula :
P(t) = A × (1 + r)^t
(1 + r) = g = Total growth percent
A = Initial population
t = time
P(t) = 8.4 million
8,400,000 = 4,200,000 × g^32
g^32 = (8400000/4200000)
g^32 = 2
Taking the root of 32 on both sides
g = 1.02189714865
g = (1 + r)
1.02189714865 = 1 + r
r = 1.02189714865 - 1
r = 0.02189714865
.rate = 0.02189714865 * 100
= 2.18971486541%
= 2.2% ( nearest tenth)
this is an example of linear equations is one variable
Answer:
50 miles per hour
Step-by-step explanation:
50 miles matches up with 1 hour, 100 miles matches up with 2 hours, and so on.
Hope this helps! :)
<span>it depends how the interest is calculated, but there's not much of a difference
assuming its continuously compouned, you use this formula: A(t)=Pe^(rt), where A is the final amount, P is the initial investment, r is the interest, and t is the time in years
you want to find t such that A(t)=18,600 so 18,600=1000e^(.0675t)
you need to use logarithm to figure it out, take the natural log of both sides
the following properties will come into use:
ln(a*b)=ln(a)+ln(b)
ln(a^b)=bln(a)
ln(e)=1
taking the natural log
ln(18,600)=ln(1000e^(.0675t))
ln(18,600)=ln(1000)+ln(e^.0675t)
ln(18600)=ln(1000) + .0675t
now solve for t: t= (ln(18600)-ln(1000))/.0675
t=43.31</span>