Answer:
A
Step-by-step explanation:
Use the table of Z-score, considering it represnts the probability of data to the left of Z-value, the needed probability would be the subtraction of Z -0.3 to Z -1.32 value.
38.209% - 9.342% =29.867%
The neaerest answer is A
Answer:
- <u>It is a combination</u>
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Explanation:
This is a combination because the order does not matter: the Beagle - Collie couple is the same as the Collie - Beagle couple, so it is only counted once.
The formula for combinations is:
For five diffferent dogs to be put in pairs, the number of different pairs is:
Liters is the best. A bathtub can hold a lot of water.
<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.
</span>
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?
</span>
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?
</span><span>
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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