Set up
Let the dimes = d
Let the pennies = p
Let the quarters = q
Equations
You cannot mean that the pennies and dimes have equal numbers. That would mean that each had 21.5 members. Now could you mean that the dime and penny amount could be the same with 43 coins that total 4.00. Four dollars means that you need 40 dimes alone. It must mean that you are including quarters.
p + d + p = 43 (1)
p = d (2)
p +10d +25q = 451 (3)
Note how this last equation = was derived. You have to multiply the dimes by 10 and he quarters by 100 and the total by 100 to get the numbers all in pennies.
Put the results of 2 into 1.
2p + q = 43 (4)
You need to modify equation 3 as well.
p + 10p + 25q = 451
11p + 25q = 451 (5)
Solve the new equations
2p + q = 43 (4)
11p + 25q = 451 (5)
Multiply 4 by 25
25(2p- + q = 43)
50p + 25q = 1075 (6) Subtract (5) from (6)
<u>11p + 25q = 451
</u>39p = 624 Divide by 39
p = 624 / 39
p = 16
Since the pennies and dimes are equal there are 16 dimes
p + d + q = 43
16 + 16 + q = 43
32 + q = 43
q = 11
Check
16 + 10*16 + 11*25 = ?
16 + 160 + 275 = ?
451 = ?
Nice problem. Thanks for posting.
4000.00 x 2.0 = $80.00 per year x 2 years = $160.00 so 6 months would earn $40.00 taking it to the $200, so 2 yrs 6 mos.
Answer:
y = 6/5x
Step-by-step explanation:
1. 
2. subtract the two number on the top of the fraction bar to get 6
3. do the same for the denominator to get 5
4. your final answer is 
The expected length of code for one encoded symbol is
where 
is the probability of picking the letter 
, and 
is the length of code needed to encode . is given to us, and we have
so that we expect a contribution of
bits to the code per encoded letter. For a string of length 
, we would then expect ![E[L]=1.375n](https://tex.z-dn.net/?f=E%5BL%5D%3D1.375n)
.
By definition of variance, we have
![\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3DE%5Cleft%5B%28L-E%5BL%5D%29%5E2%5Cright%5D%3DE%5BL%5E2%5D-E%5BL%5D%5E2)
For a string consisting of one letter, we have
so that the variance for the length such a string is
"squared" bits per encoded letter. For a string of length , we would get ![\mathrm{Var}[L]=1.859n](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3D1.859n)
.
Answer:
6 dollars
Step-by-step explanation:
