-7(2y+12)-6y=156 how am I supposed to solve this and figure out what hat “y” is?
1 answer:
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If Jerry contributes at the beginning of the month and withdraws at the end of the month, the final contribution earns 1 month's interest. The one before that earns 2 months' interest, so has a value of (1+0.017/12) times that of the last payment. In short, the sum is that of a geometric sequence with first term
a₁ = 300*(1+0.017/12)
and common ratio
r = 1+0.017/12
We assume Jerry contributes each month for 15 years, so a total of 180 payments. The sum is given by the formula for the sum of a geometric sequence.
Filling in your numbers, this is
If Jerry's contributions and withdrawal are at the end of the month, this balance is reduced by 1 month's interest, so is $61,460.
_____
We suppose the expected choice is $61,960. This supposition comes from the fact that a handwritten 4 is often confused with a handwritten 9. The usual simple calculation of future value uses end-of-the-month contributions by default. (a₁ = 300)
a₁ = 300*(1+0.017/12)
and common ratio
r = 1+0.017/12
We assume Jerry contributes each month for 15 years, so a total of 180 payments. The sum is given by the formula for the sum of a geometric sequence.
Filling in your numbers, this is
If Jerry's contributions and withdrawal are at the end of the month, this balance is reduced by 1 month's interest, so is $61,460.
_____
We suppose the expected choice is $61,960. This supposition comes from the fact that a handwritten 4 is often confused with a handwritten 9. The usual simple calculation of future value uses end-of-the-month contributions by default. (a₁ = 300)
Answer:
i) There are 40320 possible orders
ii) There are 336 possible orders for the first 3 positions.
Step-by-step explanation:
Given: The number of finalists = 8
The number of boys = 3
The number of girls = 5
To find the number of sample point the sample space S for the number of possible orders, we need to find factorial of 8!
The number of possible orders = 8!
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8
= 40320
ii) From all 8 finalist, we need to choose first 3 position. Here the order is important. So we use permutation.
nPr =
Here n = 8 and r = 3
Plug in n =8 and r = 3 in the above formula, we get
8P3 =
=
= 6.7.8
= 336
So there are 336 possible orders for the first 3 positions.