Answer:
x = 7
Step-by-step explanation:
Please give me brainliest :)
Answer:
Using <u>backward reasoning</u> we want to show that <em>"We can never get nine 0's"</em>.
Step-by-step explanation:
Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.
If we consider two cases for the second-to-last step:
<u>There were 9 </u><u>0's</u><u>:</u>
We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.
<u>There were 9 </u><u>1's</u><u>:</u>
A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.
To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.
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Angles 6 and 7, because they are alternate along the transversal.
Also angles 1 and 4.
Team A) 45 people
Team B) 55 people
A)There are two ways to solve this problem, finding the number of combinations possible for Team B, or the number of combinations possible for Team A.
Team A
It's a given that 20 mathematicians are on team A, which leavs the other 25 people for team A to be chosen from a pool of 80 (100- 20 mathletes)
80-C-25 = 80! / (25!/(80-25)!) =<span>363,413,731,121,503,794,368
</span>or 3.63 x 10^20
Solving using Team B
Same concept, but choosing 55 from a pool of 80 (mathletes excluded)
80-C-25 = 80! / (55!(80-55!) = 363,413,731,121,503,794,368
or 3.63 x 10^20
As you can, we get the same answer for both.
B)
If none of the mathematicians are on team A, then we exclude the 20 and choose 45:
80-C-45 = 80! / (45!(80-45)!) = <span>5,790,061,984,745,3606,481,440
or 5.79 x 10^22
Note that, if you solve from the perspective of Team B (80-C-35), you get the same answer</span>
