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Dmitry_Shevchenko [17]
3 years ago
10

‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️I need help ASAP 30 points and brainliest to correct answer and most helpful. Please elaborate and explain y

our answer. The chart for number 3 is attached above
In this experiment, you will be using two coins as a simulation for a real-world compound event.



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?


2. Describe how to use two coins to simulate which two pets the family has.


3. Flip both coins 50 times and record your data in a table like the one below. ( the chart is the picture that I attached)



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?


4. If the family has three pets, what is the theoretical probability that they have three dogs or three cats?


5. How could you change the simulation to generate data for three pets?

Mathematics
2 answers:
Strike441 [17]3 years ago
5 0
<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

</span>
Tems11 [23]3 years ago
3 0

I just wrote this answer so the other guy could get brainliest! He got me 100% too!!!

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Drag an answer to each box to complete this paragraph proof.
Alexxx [7]

1st box:

m<A + m<B + m<C = 180


2nd box:

substitution property


3rd box:

division property of equality


Hope it helps.

8 0
4 years ago
Read 2 more answers
A small plane flew 808 miles in 4 hours with the wind. Then on the return trip flying against the wind it travels only 488 miles
Blababa [14]

Answer:

Step-by-step explanation:

let wind velocity=x

speed of plane=y

(x+y)*4=808

x+y=808/4=202   ...(1)

(y-x)*4=488

y-x=122 ...(2)

add (1) and (2)

2y=202+122=324

y=324/2=162

from (1)

x+162=202

x=202-162=40

wind velocity=40 m/hr

speed of plane=162 m/hr

8 0
2 years ago
A=5 and b=-5 , 2ab-a^(2)
luda_lava [24]

Answer:

Step-by-step explanation:

2(5)(-5)-(5)^2

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7 0
3 years ago
Can y’all please help me question 13-14 ?!
slamgirl [31]
13. A=(-1,3) 14.m= (-4,1)
8 0
3 years ago
Please help with this and explain how to do it, I'll give 15 points
MrRissso [65]

Answer:

Option (4).

Step-by-step explanation:

Outer diameter of the hollow metallic ball = 10 centimeters

Outer radius of this ball = \frac{10}{2} = 5 cm

Volume of the outer ball V_{1} = \frac{4}{3}\pi (r)^3

                                         = \frac{4}{3}\pi (5)^3

Inner radius of the hollow metallic ball = (5 - 1) = 4 cm

Volume of the inner hollow ball V_{2} = \frac{4}{3}\pi (4)^3

Volume of the metal used in the metallic ball = V_{1}-V_{2}

                                                                           = \frac{4}{3}\pi (5)^3-\frac{4}{3}\pi (4)^{3}

                                                                           = \frac{4}{3}\pi [(5)^3-(4)^3]

Therefore, expression given in option (4) will be used to measure the volume of the hollow metallic ball.

8 0
3 years ago
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