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Anestetic [448]
3 years ago
10

In each situation, write a recurrence relation, including base case(s), for the given function. briefly explain in words why thi

s recurrence describes the situation. you do not need to solve the recurrence (that is, get a closed-form formula). (a) in a round-robin thumb-wrestling tournament with n people, everybody thumb-wrestles with everybody else. let t(n) be the total number of thumb-wrestling matches taking place among the n people. write a recurrence for t(n).
Mathematics
1 answer:
vova2212 [387]3 years ago
6 0
A thumb-wrestling match requires two thumbs, so we can either suppose that we need at least two people (n\ge2), or allow one to thumb-wrestle one's self. In either case, we'd have t(1)=t(2)=1, so let's just say we need a minimum of two players.

If we add one more person to the set of players, then the first two people would need to play 1 additional match each. So t(3)=t(2)+2.

If we add one more person, then the first three people would again each have to play 1 more match with the new person. So t(4)=t(3)+3.

And so on, so that in general, the number of games needed for everyone to play exactly one match with everyone else is given recursively by

\begin{cases}t(2)=1\\t(n+1)=t(n)+n&\text{for }n\ge2\end{cases}
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Answer:

168 units cubed for area

D is your answer

Step-by-step explanation:A=L*W*H

4*3=12*14=168 units cubed

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123*11=1353

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3 years ago
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Find an equation for the nth term of the arithmetic sequence.
babymother [125]

Answer:

The common difference would be calculated as:

(a21-a19)/2

(-164--(-58 ))/2 (Replacing the values )

(-164 +58)/2 (Changing signs)

(-106)/2(Subtracting)

-53

Then we are going to replace the common difference(d) in the sequence formal with the 21st term . It is done for finding the first term of the sequence.

a21=a0+d*(n-1)

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7 0
2 years ago
I will give you brianly look i want to know a+b=??
mr Goodwill [35]

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Step-by-step explanation:

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3 years ago
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A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds tha
ollegr [7]

Answer:

99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Step-by-step explanation:

We are given that a high school principal wishes to estimate how well his students are doing in math.

Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.

Firstly, the pivotal quantity for 99% 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 of students received a passing grade = 77%

           n = sample of tests = 40

           p = population proportion

<em>Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%

                                           level of significance are -2.5758 & 2.5758}  

P(-2.5758 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<u>99% confidence interval for p</u> = [\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

 = [ 0.77-2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } } , 0.77+2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } } ]

 = [0.5986 , 0.9414]

Therefore, 99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Lower bound of interval = 0.5986

Upper bound of interval = 0.9414

6 0
3 years ago
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