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Brrunno [24]
1 year ago
10

In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20

winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3, 4, or 5 of the 20 winning numbers. (Round all answers to the nearest hundredth of a percent.) what is the percent chance of winning
Mathematics
1 answer:
DiKsa [7]1 year ago
5 0

The percentage of winning if 20 winning numbers are randomly selected from numbers 1 to 80 is 12.48%.

Given There are 80 numbers from which 20 numbers are selected randomly in a game of Keno by player and after selection of 20 winning numbers again 20 numbers are selected and player wins if the player has correctly selected 3,4,or 5 of the 20 winning numbers.

Probability is the chance of happening an event among all the events possible.

Probability of winning is calculated as under:

Probability=C(20,3)C(60,17)/C(80,20)

We have taken C(20,3) because player will win if 5 numbers are selected from 20 winning numbers.

We have taken C(60,17) because after selection of 20 numbers 60 numbers will left from 80 and after winning from 3 numbers 17  numbers left and from total 80, 20 numbers are taken.

=C(20,3)C(60,17)/C(80,20)

=(20!/3!17!  * 60!/17!43!)/80!/20!60!

=1140*387221678682300/35353161422121743200

=12.48%

Hence the probability of winning is 12.48%.

Learn more about probability at brainly.com/question/24756209

#SPJ4

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Suppose you are working in an insurance company as a statistician. Your manager asked you to check police records of car acciden
pochemuha

Answer:

(a) 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We conclude that the the percentage of teenagers has not changed since you join the company.

(d) We conclude that the the percentage of teenagers has changed since you join the company.

Step-by-step explanation:

We are given that your manager asked you to check police records of car accidents and out of 576 accidents you selected randomly, teenagers were at the wheel in 120 of them.

(a) Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                        P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}  ~ N(0,1)

where, \hat p = sample proportion teenage drivers = \frac{120}{576} = 0.21

           n = sample of accidents = 576

           p = population percentage of all car accidents

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population population, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                     of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }} < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} ) = 0.95

<u>95% confidence interval for p</u> = [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}]

  = [ 0.21-1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }} , 0.21+1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }} ]

  = [0.177 , 0.243]

Therefore, 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We are also provided that before you were hired in the company, the percentage of teenagers who where involved in car accidents was 18%.

The manager wants to see if the percentage of teenagers has changed since you join the company.

<u><em>Let p = percentage of teenagers who where involved in car accidents</em></u>

So, Null Hypothesis, H_0 : p = 18%    {means that the percentage of teenagers has not changed since you join the company}

Alternate Hypothesis, H_A : p \neq 18%    {means that the percentage of teenagers has changed since you join the company}

The test statistics that will be used here is <u>One-sample z proportion statistics</u>;

                              T.S.  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}  ~ N(0,1)

where, \hat p = sample proportion teenage drivers = \frac{120}{576} = 0.21

           n = sample of accidents = 576

So, <u><em>test statistics</em></u>  =  \frac{0.21-0.18}{\sqrt{\frac{0.21(1-0.21)}{576} }}  

                              =  1.768

The value of the sample test statistics is 1.768.

Now at 0.05 significance level, the z table gives critical value of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the the percentage of teenagers has not changed since you join the company.

(d) Now at 0.1 significance level, the z table gives critical value of -1.6449 and 1.6449 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the the percentage of teenagers has changed since you join the company.

4 0
3 years ago
What is the GCF of (8x-6) <br><br>sos T-T
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Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to find the G.C.F. of 8x-6.

\triangle~\fbox{\bf{KEY:}}

  • G.C.F. stands for Greatest Common Factor.

So what is the G.C.F. of 8x-6? We can list all of their common factors and then select the greatest one, like so:

\longmapsto\bf{Factors\;of\;8x:} 1, 2, 4, 8, x

\longmapsto\bf{Factors\;of\;-6:} -1, 1, 2, -3, 3, -6, 6

So the GCF is:

2

Now, we can also factor it out by placing it outside the parentheses ():

2(4x-3)

See, we factored it out by dividing both terms of the expression by the G.C.F.

\ddot\bigstar Remember this...

\fbox{\bf{We\;factor\;out\;the\;G.C.F.\;by\;dividing\;all\;the\;terms\;of\;the\;expression\;by\;it}}

Hope this helps you out! :D

Ask in comments if any queries arise.

~Just a smiley person helping fellow students :)

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C

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The same x value cannot be used more than once or it isn't a function.

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