Answer:
Suppose that a couple invested $50,000 in an account when their child was born, to prepare for the child's college education. If the average interest rate is 4.4% compounded annually, ( A ) Give an exponential model for the situation, and ( B ) Will the money be doubled by the time the child turns 18 years old?
( A ) First picture signifies the growth of money per year.
( B ) Yes, the money will be doubled as it's maturity would be $108,537.29.
a = p(1 + \frac{r}{n} ) {}^{nt}a=p(1+
n
r
)
nt
a = 50.000.00(1 + \frac{0.044}{1} ) {}^{(1)(18)}a=50.000.00(1+
1
0.044
)
(1)(18)
a = 50.000.00(1 + 0.044) {}^{(1)(18)}a=50.000.00(1+0.044)
(1)(18)
a = 50.000.00(1.044) {}^{(18)}a=50.000.00(1.044)
(18)
50,000.00 ( 2.17074583287910578440507440 it did not round off as the exact decimal is needed.
a = 108.537.29a=108.537.29
Step-by-step explanation:
Hope This Help you!!
Step one: divide both sides by -2.3 to get X by itself: X=-.2. :)
Number of weekend minutes used: x
Number of weekday minutes used: y
This month Nick was billed for 643 minutes:
(1) x+y=643
The charge for these minutes was $35.44
Telephone company charges $0.04 per minute for weekend calls (x)
and $0.08 per minute for calls made on weekdays (y)
(2) 0.04x+0.08y=35.44
We have a system of 2 equations and 2 unkowns:
(1) x+y=643
(2) 0.04x+0.08y=35.44
Using the method of substitution
Isolating x from the first equation:
(1) x+y-y=643-y
(3) x=643-y
Replacing x by 643-y in the second equation
(2) 0.04x+0.08y=35.44
0.04(643-y)+0.08y=35.44
25.72-0.04y+0.08y=35.44
0.04y+25.72=35.44
Solving for y:
0.04y+25.72-25.72=35.44-25.72
0.04y=9.72
Dividing both sides of the equation by 0.04:
0.04y/0.04=9.72/0.04
y=243
Replacing y by 243 in the equation (3)
(3) x=643-y
x=643-243
x=400
Answers:
The number of weekends minutes used was 400
The number of weekdays minutes used was 243
