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arsen [322]
3 years ago
6

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?
Describe how to use two different coins to simulate which two pets the family has.
Flip both coins 50 times and record your data in a table like the one below.

Result Frequency
Heads, Heads
Heads, Tails
Tails, Heads
Tails, Tails
Total 50
Based on your data, what is the experimental probability that the family has two dogs or two cats?
If the family has three pets, what is the theoretical probability that they have three dogs or three cats?
How could you change the simulation to generate data for three pets?
Mathematics
1 answer:
olga2289 [7]3 years ago
8 0

The theoretical probability that the family has two dogs or two cats is 1/3.

(branliest will be appreciated)

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babymother [125]

Option E:

The value of m that makes the inequality true is 5.

Solution:

Given inequality is 3m + 10 < 30.

Let us first simplify the expression.

3m + 10 < 30

Subtract 10 from both side of the equation.

3m < 20 – – – – (1)

<u>To find the value of m that makes the inequality true:</u>

Option A: 20

Substitute m = 20 in (1),

⇒ 3(20) < 20

⇒ 60 < 20

It is not true because 60 is greater than 20.

Option B: 30

Substitute m = 30 in (1),

⇒ 3(30) < 20

⇒ 90 < 20

It is not true because 90 is greater than 20.

Option C: 8

Substitute m = 8 in (1),

⇒ 3(8) < 20

⇒ 24 < 20

It is not true because 24 is greater than 20.

Option D: 10

Substitute m = 10 in (1),

⇒ 3(30) < 20

⇒ 90 < 20

It is not true because 90 is greater than 20.

Option E: 5

Substitute m = 5 in (1),

⇒ 3(5) < 20

⇒ 15 < 20

It is true because 15 is less than 20.

Hence the value of m that makes the inequality true is 5.

Option E is the correct answer.

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586/10 -> 293/5 -> 58 3/5
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Answer:

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Step-by-step explanation:

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