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inna [77]
3 years ago
15

Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
user100 [1]3 years ago
5 0
Flip a coin twenty five times, the purpose of this is to show that theoretical and experimental do not always overlap.

Theoretically, it should be a fifty-fifty chance.

In the experiment because you do it a odd amount of times, 25, each flip will be worth a four percent chance.

You would not be able to make a fifty fifty chance with that amount of flips.

Also here:

1.) 13 Heads, 12 tails

2.) 48% chance for the coin to land on tails, 52% chance for the coin to land on heads.

3.) The theoretical probability of a coin landing on heads is 50% of the time that the coin is flipped. This is because there are two possibilities with an equal likelihood of happening 

4) The theoretical probability and experimental probability are different as theoretically there would be an equal likelihood or probability and in the experiement, there was a higher probability for the coin to land on heads. 

vladimir1956 [14]3 years ago
5 0

Flip a coin twenty five times, the purpose of this is to show that theoretical and experimental do not always overlap.


Theoretically, it should be a fifty-fifty chance.


In the experiment because you do it a odd amount of times, 25, each flip will be worth a four percent chance.


You would not be able to make a fifty fifty chance with that amount of flips.


Also here:


1.) 13 Heads, 12 tails


2.) 48% chance for the coin to land on tails, 52% chance for the coin to land on heads.


3.) The theoretical probability of a coin landing on heads is 50% of the time that the coin is flipped. This is because there are two possibilities with an equal likelihood of happening 


4) The theoretical probability and experimental probability are different as theoretically there would be an equal likelihood or probability and in the experiement, there was a higher probability for the coin to land on heads. 

1

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6 0
3 years ago
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djyliett [7]

Answer:

i know it is C last year i took that quiz and well i studied /j

Step-by-step explanation:

6 0
2 years ago
Assume that all six outcomes of a six-sided number cube have the same probability. What is the theoretical probability of each r
Margarita [4]
Hey there! I am on the same one. :) I will help you out a little. 

<span>Assume that all six outcomes of a six-sided number cube have the same probability. What is the theoretical probability of each roll? 
• 1: 1/6
• 2: 2/6
• 3: 3/6
• 4: 4/6
• 5: 5/6
• 6: 6/6
</span>
<span>Using the uniform probability model you developed, what is the probability of rolling an even number? 
1/6 Roll a number cube 25 times. Record your results here. 

</span><span> <span><span> <span> <span>1st toss=</span>6</span> </span> <span> <span> <span>2nd toss=</span>4</span> </span> <span> <span> <span>3rd toss=</span>6</span> </span> <span> <span> <span>4th toss=</span>6</span> </span> <span> <span> <span>5th toss=</span>3</span> </span> <span> <span> <span>6th toss=</span>3</span> </span> <span> <span> <span>7th toss=</span>4</span> </span> <span> <span> <span>8th toss=</span>2</span> </span> <span> <span> <span>9th toss=</span>6</span> </span> <span> <span> <span>10th toss=</span>5</span> </span> <span> <span> <span>11th toss=</span>1</span> </span> <span> <span> <span>12th toss=</span>4</span> </span> <span> <span> <span>13th toss = </span>5</span> </span> <span> <span> <span>14th toss =</span>1</span> </span> <span> <span> <span>15th toss=</span>4</span> </span> <span> <span> <span>16th toss=</span>2</span> </span> <span> <span> <span>17th toss=</span>2</span> </span> <span> <span> <span>18th toss=</span>2</span> </span> <span> <span> <span>19th toss=</span>6</span> </span> <span> <span> <span>20th toss=</span>5</span> </span> <span> <span> <span>21st toss=</span>3</span> </span> <span> <span> <span>22nd toss=</span>4</span> </span> <span> <span> <span>23rd toss=</span>3</span> </span> <span> <span> <span>24th toss=</span>3</span> </span> <span> <span> 25 toss=5
How many results of 1 did you have? __2____________ How many results of 2 did you have? ____4__________ How many results of 3 did you have? ____5__________ How many results of 4 did you have? ______5________ How many results of 5 did you have? ______4________  How many results of 6 did you have? ______5________
Based on your data, what is the experimental probability of each roll?  <span> 1. 2/25 or 0.08 

2. 4/25 or 0.16

3. 5/25 or 0.24

4. 5/25 or 0.2

5.4/25 or 0.16
<span>
6. 5/25 or 0.2</span></span>Using the probability model based on observed frequencies, what is the probability of rolling an even number? 3/6 = ½ or 0.5
Was your experimental probability different than your theoretical probability? Why or why not? <span>It somewhat is! The denominator is 25 for the experimental probability, and 6 for the theoretical probability.</span><span>
</span><span>Have a lovely day! Cheerio. :) </span></span> </span> </span></span>
8 0
3 years ago
Read 2 more answers
Jenny runs a daycare center and charges $60 per day for each child. There are currently 50 children enrolled at the daycare. At
Goshia [24]

Answer:

(60 - 4x)(50 + 2x) = 2,800

Step-by-step explanation:

5 0
3 years ago
Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p
Ymorist [56]

Answer:

E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours

Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

A\sim Uniform(1 ,7)

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

B\sim Exp(\lambda=\frac{1}{30 min})

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

E(A +B)

Since the two random variables are assumed independent, then we have this

E(A+B) = E(A)+E(B)

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by E(X) =\frac{a+b}{2} where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

E(A)=\frac{1+7}{2}=4 hours

The expected value for the exponential distirbution is given by :

E(X)= \int_{0}^\infty x \lambda e^{-\lambda x} dx

If we use the substitution y=\lambda x we have this:

E(X)=\frac{1}{\lambda} \int_{0}^\infty y e^{-\lambda y} dy =\frac{1}{\lambda}

Where X represent the random variable and \lambda the parameter. If we apply this formula to our case we got:

E(B) =\frac{1}{\lambda}=\frac{1}{\frac{1}{30}}=30min

We can convert this into hours and we got E(B) =0.5 hours, and then we can find:

E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours

And in order to find the variance for the random variable A+B we can find the individual variances:

Var(A)= \frac{(b-a)^2}{12}=\frac{(7-1)^2}{12}=3 hours^2

Var(B) =\frac{1}{\lambda^2}=\frac{1}{(\frac{1}{30})^2}=900 min^2 x\frac{1hr^2}{3600 min^2}=0.25 hours^2

We have the following property:

Var(X+Y)= Var(X)+Var(Y) +2 Cov(X,Y)

Since we have independnet variable the Cov(A,B)=0, so then:

Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2

3 0
3 years ago
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