solution
this is what I did. total number of ways without any restriction =900. we have to subtract the case when at least one child doesn't get any fruit. let that be C1∪C2∪C3 and we use the principle of mutual inclusion and exclusion to find that. C1∪C2∪C3=C1+C2+C3−(all two intersections)+3 intersections (no one gets a fruit case) =3⋅60−3+1=178 so case where each child gets at least one fruit =900−178=722.
Is this right, someone saying answer =49
Answer:
- see the attachment for a graph
- yes, you can carry 5 math books in one load (along with 0–2 science books)
Step-by-step explanation:
If x and y represent the number of math and science books you're carrying, respectively, then 3x and 4y represent their weights in pounds.
The total weight of the carry will be 3x+4y, and you want that to be at most 24 pounds. The expression modeling this is ...
... 3x +4y ≤ 24
A graph of this inequality is shown in the attachment. (We have added the constraints that the number of books not be negative.)
___
5 math books will weigh 5·3 = 15 pounds, so will be within the limit you can carry.
Answer:
i. 33
ii. 1488
iii. 65
Step-by-step explanation:
i. 4x² − 3y² + 5z²
Putting the values x=2, y=-1 and z=2
4(2)² − 3(-1)² + 5(2)²
4×4 − 3×(1)² + 5×(4)
(16) - (3×1) + (20)
16 - 3+ 20
36 - 3
33
ii. 3x³ − 2x(4yz+5x²)
Putting the values x=2, y=-1 and z=2
[3(2)³ − 2×2(4×(-1)×2 + 5×(2)²)]
[3(8)³ − 2×2(-8 + 5×4)]
[3 ×512 - 2×2(-8 +20)]
[1536 - 4(12)]
[1536 - 48]
1488
iii. 1−4x(yz+3xy)
Putting the values x=2, y=-1 and z=2
[1 - 4 × 2((-1)×2 + 3×2×(-1))]
[1- 8(-2 + (-6))]
[1- 8 (-2 -6)]
[1 - 8(-8)]
[1 +64]
65
Answer:
B is the answer :)
Step-by-step explanation:
25% of the results are inbetween each of the areas. They're called quartiles :)
