If $10,000 is invested in an account at 4.8% annual simple interest, how long will it take for the account balance to grow to $11,200?
i am confused. how do you do this?

First i would like to thank you for asking this, im reviewing for my math final and forgot to study this!! and im giving the equation from my textbook modified for your problem so 

the formula is A= P(1+ r/n)^nt
Balance= A (in your case would be 11,200) the money you have now
Principle=P= (in your case 10,000) the money you invest
Time=t= (unknown in your case) the amount of time
Compoundings per year=n= (1 in your case since it is annually)
annual interest rate=r= (in your case 4.8% but you want to put it in decimal form which would be .048) 

plug in your numbers to get your equation of 

11200=10000(1+ .048/1)^1(t) 
then we solve for t
start by dividing both sides by 10000

1.12=(1+.048)^t (i simplified everything multiplied or divided by 1 because it would be the same) 

then add everything in the parentheses
1.12=(1.048)^t
 
then we have to use logarithms... i know super fun, i hate them too... anyway, take the log of both sides

log(subscript 10) (1.12)= log(subscript 10) (1.048^t)
we can then move the t to the other side of the log we will pretend x means times instead of it being a variable

log(subscript 10) (1.12)= t x log(subscript 10) (1.048)
we then want to isolate the term by dividing both sides by the constant number of log(subscript 10) (1.048) to get

(log(subscript 10) (1.12))/ (log(subscript 10) (1.048) =t
and then calculate in your special fancy high school calculator they forced us to buy to use like twice...to get your answer of 

2.417235865 years and you  can round that as you need

remember to always double check your work.. for this i am going to round to 2.42 years 
to check, go back to your original formula and plug our answer into t and make sure the two sides equal each other

11200=10000(1.048)^2.42
this comes out to 11200=11201.45153 but that is because we rounded, without rounding it is 11200=11200 so it is the right answer

i hope this helped!!!

 

The mode is the most frequently occurring value.