Answer:
D
Step-by-step explanation:
You cannot divide a decimal by a full number and also you move the decimal too places too the right so the bottom option
2(3x+4y) = 2(24)
6x+8y = 48
Add both the equations
6x + 8y = 48
+ x - 8y = -20
—————————
7x = 28
x = 28/7
x = 4
Plug in this value in the second equation
(4) - 8y = -20
-8y = -24
y = -24/-8
y = 3
Therefore:
x = 4
y = 3
Answer:
1120 possible ways
Step-by-step explanation:
In order to find the answer we need to be sure what equation we need to use.
From the given example, let's consider initially only men. Because you have a total of 8 men and we need to chose only 3 men, let's suppose that the 3 chosen men are A, B, and C.
Because A,B,C is the same as choosing C,B,A, which means it doesn't matter the order of the chosen men, we need to use a 'combination equation'.
Because we have two groups (women and men) then we have:
Possible ways = 8C3 * 6C3 (which are the combinations for women and men respectively). Remember that:
nCk=n!/((n-k)!*k!) so:
Possible ways = 8!/((8-3)!*3!) * 6!/((6-3)!*3!) = 56* 20 = 1120.
In conclusion, there are 1120 possible ways.
To answer the question, you need to determine the amount Mr. Traeger has left to spend, then find the maximum number of outfits that will cost less than that remaining amount.
Spent so far:
... 273.98 + 3×7.23 +42.36 = 338.03
Remaining available funds:
... 500.00 -338.03 = 161.97
The cycling outfits are about $80 (slightly less), and this amount is about $160 (slightly more), which is 2 × $80.
Mr. Traeger can buy two (2) cycling outfits with the remaining money.
_____
The remaining money is 161.97/78.12 = 2.0733 times the cost of a cycling outfit. We're sure he has no interest in purchasing a fraction of an outfit, so he can afford to buy 2 outfits.