Answer:
15 boxes
Step-by-step explanation:
Because 71 divided by 5 equals to 14.2, you know that you need at least 14 boxes. However, you cannot have 14.2 boxes, because that implies that you have a fifth of a box, which cannot happen. But because each box can only hold up to 5 cards max, you need to get a 15th box for the last card.
-18+54
-17+53
-16+52
-15+51
-14+50
-13+49
-12+48
-11+47
-10+46
-9+45
-8+44
-7+43
-6+42
-5+41
-4+40
-3+39
-2+38
-1+37
1+35
2+34
3+33
4+32
5+31
6+30
7+29
8+28
9+27
10+26
11+25
12+24
13+23
14+22
15+21
16+20
17+19
18+18
Pet show has stations for each animal. There are 5 rabbits, 7 cats, 8 dogs, and 4 hamsters.
1. How many cats need to be added to make the probability 1/2.
For the probability to be 1/2, we need there to be 1 cat for every 2 animals. There are 5+8+4 = 17 other animals besides cats so we need to increase the amount of cats to 17 so that it is 1/2 of the total. We already have 7 so we need 17-7 = 10 more cats.
2. Assume we added the cats. What's the probability of picking a dog?
Since we added the cats, the total number of animals is now 17+17 = 34
There are 8 dogs, so the probability of picking a dog will be 8/34 which will reduce to 8/34 = 4/17 (you will pick 4 dogs every 17 attempts).
3. No more animals added. What is the probability of picking a rabbit or hamster?
Total rabbits and hamsters is 5+4 = 9 so there is a 9/34 chance of picking a rabbit or hamster. (you will pick a rabbit or hamster 9/34 attempts).
4. What is the probability of NOT picking a goldfish?
Be careful here... the question is what is the probability that you won't pick a goldfish? Well there are no goldfish, so there is a 34/34 (100%) chance that you won't pick a goldfish. You will NOT pick a goldfish 34 out of 34 times.
21x³ - 18x²y + 24xy²
Simplify!
7x³ - 6x²y + 8xy²
Answer:
D
Step-by-step explanation:
The correct answer choice is D. Note how BC and MN appear to have nearly the same lengths, whereas none of the other statements are true. Note, also, that this diagram is not to scale; you could make the problem easier by redrawing MNOP so that it has the same orientation as does ABCD and the same corresponding side lengths.
