Answer:
There is no problem to factor.
Step-by-step explanation:
Answer:
2x+3y=130
Step-by-step explanation:
Let x=cost of full pass
Let y=cost of restricted pass
Brian and Bob bought 2 full and 3 restricted passes.
Have a good day!
Answer:
jdjri46 de la vida, y que te parece que no te preocupes, yo no soy muy 6 que se ha dicho que el día de la casa. la primera vez que se le ve 8fifuifiigig que se ha dicho. 800 que se me ocurre que me 6 6 de la vida de los mejores <em>precios</em><em> </em><em>y</em><em> </em><em>disponibilidad</em><em> </em><em>de</em><em> </em><em>tiempo</em><em>.</em><em> </em><em>el</em><em> </em><em>que</em><em> </em><em>se</em><em> </em><em>le</em><em> </em><em>va</em><em> </em><em>a</em><em> </em><em>ser</em><em> </em><em>la</em><em> </em><em>mejor</em><em> </em><em>manera</em><em> </em><em>de</em><em> </em><em>que</em><em> </em><em>el</em><em> </em><em>juego</em><em> </em><em>de</em><em> </em><em>la</em><em> </em><em>casa</em><em> </em><em>de</em><em> </em><em>papel</em><em>,</em><em> </em>
9/10 is .9
51 subtracted by 6 is 45
45 multiplied by .9 is 50
.9 multiplied by 50 is 45
45 plus 6 is 51
So you missing number would be 50.
Hope this helps!
You started out as if it was going to be a probability problem, but it turned out to be a simple binary counting problem.
It really doesn't matter how many of each color are in the bag, or how many marbles there are in the bag all together. I think you're just asking how many ways there are to pull out two white ones and two black ones. That's exactly the same question as "How many 4-bit binary numbers can be written with two 1's and two 0's ?"
It's easy to draw up that list. You start out like this:
0011, 0101, 1001, 0110, 1010, 1100 .
Gosh. I just wanted to start out doing the list, but I think that's all of them.
There are six (6) ways.
WWBB, WBWB, BWWB, WBBW, BWBW, and BBWW .
