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!!
Grab some paper, a pencil, and a ruler. Make a 6 by 6 square
X = approximately 633
Steps:
lnx + ln3x = 14
ln3x^2 = 14 : Use the log property of addition which is to multiply same log together so you multiply x and 3x because they have log in common
(ln3x^2) = (14) : take base of e on both sides to get rid of the log
e e
3x^2 = e^14 : e cancels out log on the left side and the right side is e^14
x^2 = e^14 / 3 : divide both sides by 3
√x^2 = <span>√(e^14 / 3) : take square root on both sides to get rid of the square 2 on x
</span>
x = √(e^14) / <span>√3 : square root cancels out square 2 leaving x by itself
x = e^7 / </span>√3 : simplify the √(e^14) so 14 (e^14) divide by 2 (square root) = 7<span>
x = </span>633.141449221 : solve
43 to 49 is 6
49 to 61 is 12
61 to 79 is 18
add 6 to each
79=4th row
79+18+6=79+24=103
103+24+6=103+30=133
133+30+6=133+36=169
7th row=169 seats
