Hi can someone help me for correcting my text
Your answer is A have a good day
<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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Answer:
number of ways = 120
Step-by-step explanation:
The number of ways five children can pose for a photograph line up in a row is given by the number of permutations of 5 elements in 5 different positions (positions in the line), then
number of ways = number of permutations of 5 elements = 5! = 5 * 4 * 3 * 2 * 1 = 120
Since the first children that occupies the line can be on any of the positions (5 positions) , but then the second one can choose any of the 4 remaining positions (since the first children had already occupied one) , the third can choose 3 ... and so on.
