The answer is 130 degrees b
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!!
The answer choices are sufficiently far apart that you can work this backward. The sum will be ...
236,196*(1 + 1/3 + 1/9 + 1/27 + ...)
so a reasonable estimate can be given by an infinite series with a common ratio of 1/3. That sum is
236,196*(1/(1 - 1/3)) = 236,196*(3/2)
Without doing any detailed calculation, you know the best answer choice is ...
354,292
_____
There are log(236196/4)/log(3) + 1 = 11 terms* in the series, so the sum will be found to be 4(3^11 -1)/(3-1) = 2*(3^11-1) = 354,292.
Using the above approach (working backward from the last term), the sum will be 236,196*(1-(1/3)^11)/(1-(1/3)) = 236,196*1.49999153246 = 354,292
___
* If you just compute log(236196/4)/log(3) = 10 terms, then your sum comes out 118,096--a tempting choice. However, you must realize that the last term is larger than this, so this will not be the sum. (In fact, the sum is this value added to the last term.)
Answer:
I think it is B
Step-by-step explanation:
Answer:
0.67450852
Step-by-step explanation:
you only use your calculator and you put the tan 34 and gives you the answer
