I think no ! Hope this helps ..... what class is this for
The effective annual interest rate is:
i = (1 + 0.064/12)^12 - 1 = 0.066
In year 1: the interest is $613.80 (multiple $9300 by 0.066)
In year 2: the interest is $654.31 (add interest from year 1 to $9300 and multiply by 0.066)
In year 3: the interest is $656.98 (do the same as year 2)
In year 4: the interest is $657.16
The total interest is: $2582.25
The present worth of this amount is:
P = 2582.23 / (1 + 0.066)^4 = $1999.72
The answer is $1999.72.
Answer:
y=2x-3
Step-by-step explanation:
show work
1=(2*2)+b
1=4+b
1-4=-3
-3=b
check work
y=2x-3
y=(2*2)-3
y=4-3
y=1
Answer:
the answer is 2.25
Step-by-step explanation:
you divide 6.75÷ 3
Might have to experiment a bit to choose the right answer.
In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.
Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456
In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.
Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.
