Weekly Reflection 9-7-18

Exponential Word Problems
I will be able to use compound interest to solve exponential problems.

One example of an Exponential word problem is:
Find the accumulated amount of money after 5 years if $4,300 is invested at 6% per year compounded quarterly. Find the growth factor. 
In order to solve this problem you have to find out the accumulated amount after (t) number of years, which in this problem is what we are supposed to find out, the principle (original amount) which is 4300(P), the nominal interest rate per year, which is 6% or .06(r), the number of periods or years(n), which is 4 because the money is compounded quarterly which is 4 times a year, and the number of years(t), which is 5. Then you plug these numbers into the compound interest formula. A=P(1+r/n)^n*t ------->  A=4300(1+.06/4)^4*5 --------->  A=5791.48
The accumulated amount after 5 years is approximately $5791.48. The growth factor is 1.015

One misconception I had wasn't with exponential word problems but with the exponential to log homework on one problem I added ln10=x-1 and got ln11=x but I wasnt supposed to add because they are 2 separate numbers the answer was ln10+1=x
I overcame my misconception when Mrs. Burton explained that this was incorrect.

EQ: How much will my dream car the 2018 Wrangler JK Golden Eagle 4x4 ($34,675) be worth in 5 years  at 15%  continuously compounded interest?
A=Pe^rt ----1--> A=34,675e^.15*5 --------> A=73,406.98
The value of the car in 5 years will be approximately $73,406.98 

Comments

Post a Comment

Popular posts from this blog

Weekly Reflection 11-30-18

Image
Empirical Rule:  There is no objective Empirical Rule Example: When using the empirical rule the problem has to call for normal distribution or else you can  not use the empirical rule. There are three numbers you have to remember for empirical rule those numbers are; 68%, 95%, and 99.7%. These numbers represent the intervals of information that is included in these margins. The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives  16 1 6  years; the standard deviation is  1.7 1 . 7  years. Use empirical rule to estimate the probability of a gorilla living longer than 14.3 years?  In order to find   the probability I would need to make a normal distribution table and place 16 years as the mean and add and subtract 1.7 as intervals. The numbers I would use would be 9.2, 10.9, 12.6, 14.3, 16, 17.7, 19.4, 21.1, 22.8. Now that I know the numbers I can figure out the percentage. The number 14.3 ...
Read more

Weekly Reflection 11-9-18

Piecewise Word Problems: I will be able to be able to write piecewise functions Piecewise word problem example: The admission rates at an amusement park are as follows:                 Children 5 years old and under: Free                 Children between 5 years and 12 years ,inclusive: $10                 Children between 12 years and 18 years, inclusive: $25                 Adults: $35 To solve this this first thing I would do is to write down the ages. The ages will serve as the x because they change. So for the first I say 0 less or than equal to x less than or equal to 5, because you can not just say less than or equal to 5 because that would include negative numbers and no one's age is a negative number. So 0 would have to be the lowest number. For the second I  say 5 less than x less than or equal to 12. Th...
Read more

Weekly Reflection 11-16-18

Arithmetic and Geometric Sequences: I will be able to understand arithmetic series and sequence. I will be able to understand geometric series and sequence. Example of arithmetic sequence: -9,-2,5,12,___,___,___,___                        a(n)=a(1) + (n-1)d I do not know what my end term[a(1)] is so that is what I am looking for. I do know what my first term [a(1)] is, it is -9, I have 8 terms [n] and I can find my common difference by subtracting -2 from -9. which is 7. To solve all I need to do is add 7 to 12 then to that number and so on until I have all the terms I need.12+7=19, 19+7=26, 26+7= 3, and 33+7= 40. Now that I have my four terms I can add them to my series. Solution: -9,-2,5,12, 19,26,33,40 Example of geometric sequence: Find two geometric means between -2 and 54 -2, __, __, 54 a(n)=a(1)*r^(n-1) I do know what my end term is, it is 54, my first term is -2, I do not know what the common ratio is I h...
Read more