Step-by-step explanation:
I assume it is meant to draw 2 blue balls in a row out of the full pool of balls (no other draws happen before or during this process).
tricky, your teacher.
I think the temptation is to simply say 1/3×1/3 = 1/9.
but is it ?
the probabilty to draw one blue ball out of the complete system (no ball missing) is
x / (6+4+x) = 1/3
x = (6+4+x)/3
3x = 6 + 4 + x = 10 + x
2x = 10
x = 5
so, we have a system of 6 orange, 4 green and 5 blue balls (15 all together) .
the probability at our first pull to draw a blue ball is the indicated 5/15 = 1/3.
but now, for the second ball, we have only 6 orange, 4 green and 4 blue balls in the pool (14 altogether).
and so the probability to draw a blue ball out of this is
4/14 = 2/7.
therefore, the probability to draw 2 blue balls is now the combination
1/3 × 2/7 = 2/21
aha ! so, it is a little bit less than the originally suspected probability of 1/9.
First multiply every thing by 7 to get rid of the fraction.
(y/7)(7) +6(7)=11(7)
y+42=77
y=35
5 + 4x
I think that’s right
Answer:
r = 3.21
Step-by-step explanation:
Answer:
1050 g < weight ≤ 1150 g
Step-by-step explanation:
Let w represent the weight of the package in grams. The the number of 100-gram increments after the first 250 grams is given by ...
⌈(w-250)/100⌉ . . . . . . . where ⌈ ⌉ signifies the <em>ceiling</em> function
and the charges for a package exceeding 250 grams will be ...
0.65 + 0.10⌈(w -250)/100⌉ = 1.55
0.10⌈(w -250)/100⌉ = 0.90 . . . . . . . . subtract 0.65
⌈(w -250)/100⌉ = 9 . . . . . . . . . . . . . . . divide by 0.10
8 < (w-250)/100 ≤ 9 . . . . . . . . . . . . . . meaning of ceiling function
800 < w -250 ≤ 900 . . . . . . . . . . . . . multiply by 100
1050 < w ≤ 1150 . . . . . . . . . . . . . . . . . add 250
The weight in grams could be greater than 1050 and at most 1150 for a charge of $1.55.