Well, we know the numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11. From these numbers, 6 are odd. So, 6/11 is the probability that it is odd.
You take the graph of f(x) and shift it down four units. (Assuming f:R->R).
To solve this problem, simply substitute the values that have been given to you for both X and y. I believe in this question, it would be asking you to verify or check if the values of X and y are correct.
X = 1/4
Y = 1/2.
2x^2 + y = 5/8
2(1/4)^2 + 1/2 = 5/8
2(1/16) + 1/2 = 5/8
1/8 + 4/8 = 5/8
5/8 = 5/8.
1/2 = 4/8
2/16 = 1/8 in lowest terms.
You have effectively wasted 1 hour of my time. Congratulations.
115 = 65 + 50
45(65) + 35(50) = 2925 + 1750 = 4675
45(50) + 35(65) = 2250 + 2275 = 4525
115 = 55 + 60
45(55) + 35(60) = 2475 + 2100 = 4575
Therefore 55 large boxes and 60 small boxes. Absolutely no more of these questions; I got stuck for 40 minutes using trial and error, even though that's the only way you can solve two variables in one equation with integers.
