Initially, Charlotte owes $7680. She finishes her payments after a total of 6 + 36 = 42 months. Using a simple compounding formula, the amount she owes is worth P at the end of 42 months, where P is:
P = 7680 * (1 + .2045/12)^42 = 15616.67379
Now, the first installment she pays (at the end of six months) is paid 35 months in advance of the end, so it is worth x * (1 + .2375/12)^35 at the end of her loan period.
Similarly, the second installment is worth x * (1 + .2375/12)^34 at the end of the loan period.
Continuing, this way, the last installment is worth exactly x at the end of the loan period.
So, the total amount she paid equals:
x [(1 + .2375/12)^35 + (1 + .2375/12)^34 + ... + (1 + .2375/12)^0]
To calculate this, assume that 1+.2045/12 = a. Then the amount Charlotte pays is:
x (a^35 + a^34 + ... + a^0) = x (a^36 - 1)/(a - 1)
Clearly, this value must equal P, so we have:
x (a^36 - 1)/(a - 1) = P = 15616.67379
Substituting, a = 1 + .2045/12 and solving, we get
x = 317.82
Answer:
y = 12x-6
Step-by-step explanation:
y=mx+b
y = 12x-6
Answer:
- 6 x - 182
Step-by-step explanation:
He ran .45 miles per minute
Answer:
Danny will watch all of them on the same day again on August 31
Step-by-step explanation:
Here, we have a word problem.
The shows and frequency of watching are;
Show A Every 2 days
Show B Every 3rd day
Show C Every 5th day
Show D Every 10th day
Danny watched all four shows on August 1, therefore, the day he will watch all four shows can be found by finding the Lowest Common Multiple, LCM of the frequency of his watching each of the shows. That is, the LCM of 2, 3, 5 and 10
Therefore we have, since 10 is the highest frequency and 3 is not a factor of 10, then the LCM that applies to 3 and 10 is 3×10 = 30.
Since the other frequencies are also a factor of 30, then the LCM of 2, 3, 5, and 10 is 30
Therefore, Danny will watch all four shows again on the same day again in 30 days after August 1. That is August (1 + 30) = August 31.
