Answer:
Step-by-step explanation:
a.) The worst-case height of an AVL tree or red-black tree with 100,000 entries is 2 log 100, 000.
b.) A (2, 4) tree storing these same number of entries would have a worst-case height of log 100, 000.
c.) A red-black tree with 100,000 entries is 2 log 100, 000
d.) The worst-case height of T is 100,000.
e.) A binary search tree storing such a set would have a worst-case height of 100,000.
Answer: A. X(–2, 2), Y(–10, 2), Z(–6, 10)
Step-by-step explanation: This looks a bit tough, but knowing that scale factors less than 1 are ½, ¼, etcetera, I come to this answer, also knowing that in the end, they must have an unproportional relationship.
I hope this is correct, and as always, I am joyous to assist anyone at anytime.
Answer:
"13 / 204 ; No, they are dependent events
"
Step-by-step explanation:
The probability of getting a club at first throw is simply total number of clubs divided by total number of cards in the deck.
Total clubs is 13
Total cards in deck is 52
Hence, P(clubs) = ![\frac{13}{52}](https://tex.z-dn.net/?f=%5Cfrac%7B13%7D%7B52%7D)
<em>Since the first card IS NOT replaced, the deck has now 51 cards. We want probability of diamond. There are still 13 diamonds, so probability of diamond now would be 13/51</em>
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Hence, Probability that first card is a club and the second card is a diamond should be 13/52 MULTIPLIED by 13/51
P(1st Club, 2nd Diamond) = ![\frac{13}{52}*\frac{13}{51}=\frac{13}{204}](https://tex.z-dn.net/?f=%5Cfrac%7B13%7D%7B52%7D%2A%5Cfrac%7B13%7D%7B51%7D%3D%5Cfrac%7B13%7D%7B204%7D)
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<u>Note: </u>They are dependent events since we are calculating probability without replacement. The first draw affected the total number of cards in the 2nd draw.
2nd answer choice is right.
Answer:
not sure no attachments but i think its 49 or 37 or both together 88
Step-by-step explanation: