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dezoksy [38]
3 years ago
12

The boston celtics have won 16 nba championships over approximately 50 years. Thus it may seem reasonable to assume that in a gi

ven year the celtics win the title with probability p = 16/50 = 0.32, independent of any other year. Given such a model, what would be the probability of the celtics winning eight straight championships beginning in 1959?
Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0

We have been given that The boston celtics have won 16 nba championships over approximately 50 years. Therefore, it can be concluded that in any given year the probability of celtics winning the title will be p = 16/50 = 0.32

For Celtics to win eight straight championships in 1959, we will be required to multiply the probability of winning championship in one year to itself 8 times.

Therefore, the required probability is:

0.32^{8}=0.0001099511627776

Therefore, the required probability is almost 0.00011.

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LuckyWell [14K]

Answer:

Not sure sorry Im really bad at this stuff

Step-by-step explanation:

8 0
3 years ago
Is the following statement true? Explain.
sukhopar [10]

Step-by-step explanation:

We can prove the statement is false by proof of contradiction:

We know that cos0° = 1 and cos90° = 0.

Let A = 0° and B = 90°.

Left-Hand Side:

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Right-Hand Side:

cos(A) + cos(B) = cos(0°) + cos(90°)

= 1 + 0 = 1.

Since LHS =/= RHS, by proof of contradiction,

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4 0
3 years ago
I need help with finding 25 different numbers that add up to 51. It must be a six digit number. Example would be 6+7+5+8+9+6 = 4
USPshnik [31]
What you could do it is. Start adding numbers and once you get close to
6 numbers See if you are close to 51 and switch around the numbers to make it equal 51 hope this helps
6 0
3 years ago
Is the sequence 5, 9, 15, an arithmetic sequence? Explain.
igor_vitrenko [27]

Answer:

  no

Step-by-step explanation:

An arithmetic sequence has a common difference between adjacent terms. Here, the differences are ...

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The differences are different, not common, so this is NOT an arithmetic sequence.

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