<span>1. Suppose that a family has an equally likely chance of having a cat or a dog. If they have two pets, they could have 1 dog and 1 cat, they could have 2 dogs, or they could have 2 cats.
What is the theoretical probability that the family has two dogs or two cats?
25% chance
</span><span>2. Describe how to use two coins to simulate which two pets the family has.
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You could use the coins to simulate which pet the family has by flipping them and having head be dog and tails be cat (or vice-versa).
<span>3. Flip both coins 50 times and record your data in a table like the one below.
</span><span>Based on your data, what is the experimental probability that the family has two dogs or two cats?
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Based on the results, I concluded that for Heads, Heads (which could be dogs or cats) there was a 24% chance and for Tails, Tails there was a 26% chance
<span>4. If the family has three pets, what is the theoretical probability that they have three dogs or three cats?
1/8 chance (accidentally messed up there) or 12.5%
</span><span>5. How could you change the simulation to generate data for three pets?
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To flip 3 coins and add more spots on the chart.
I hope that this helps because it took a while to write out. If it does, please rate as Brainliest
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I think it's called a coordinate plane
Sorry if I'm wrong
y = 80x - 60
Answer is B. y = 80x - 60
Double check:
When x = 1, y = 80(1) - 20 = 80 -60 = 20
When x = 2, y = 80(2) - 20 = 160 -60 = 100
When x = 3, y = 80(3) - 20 = 240 -60 = 180
When x = 4, y = 80(4) - 20 = 320 -60 = 260
When x = 5, y = 80(5) - 20 = 400 -60 = 340
Answer:
2 pt + 1c
Step-by-step explanation:
There are 2 cups in a pint. So, 2c = p, dividing 2 from both sides, we get
c = (1/2)p
Replace c with 1/2 p in the equation
5 pt - (2pt + (1/2)p)
= 5pt - 2pi - (1/2)p
= 3pt - (1/2)p
= 2pt + (1/2)p
Replace back (1/2)p for c
2pt + (1/2)p =
2 pt + 1c
Y= -x^2. That’s the answer
