There are 60 seconds in a minute.
There are 60 minutes in an hour.
This means that there'd be 600 minutes in 10 hours.
And if there are 60 seconds in a minute, there are 600 x 60 seconds in 600 minutes.
600 x 60 = 600 x 10 x 6 = 6000 x 6 = 36000
So we know that there are 36000 seconds in 10 hours.
1/2r - 3 = 3(4-3/2) -----> Simplify the right side.
1/2r - 3 = 3(1/2) -----> Multiply
1/2r - 3 = 3/2 ----> Add 3 to both sides
1/2r = 6/2 + 3/2
1/2r = 9/2 -----> Divide both sides by 1/2 which is the same as multiplying my 2
r = 18/2
r = 9
Hope this helps!
If you're rolling 5 dice, and all dice have a 1 on them, then you still have a 4/5 chance that four of the five dice will show 1.
Answer:
Maximize C =


and x ≥ 0, y ≥ 0
Plot the lines on graph




So, boundary points of feasible region are (0,1.7) , (2.125,0) and (0,0)
Substitute the points in Maximize C
At (0,1.7)
Maximize C =
Maximize C =
At (2.125,0)
Maximize C =
Maximize C =
At (0,0)
Maximize C =
Maximize C =
So, Maximum value is attained at (2.125,0)
So, the optimal value of x is 2.125
The optimal value of y is 0
The maximum value of the objective function is 19.125