A and D are the correct answers. They are on opposite sides of the transversal and are on the outside of the parallel lines.<span />
(a)
since 13 is prime.
(b)
, and there are 81/3 = 27 multiples of 3 between 1 and 81, which leaves 81 - 27 = 54 numbers between 1 and 81 that are coprime to 81, so
.
(c)
; there are 50 multiples of 2, and 20 multiples of 5, between 1 and 100; 10 of these are counted twice (the multiples of 2*5=10), so a total of 50 + 20 - 10 = 60 distinct numbers not coprime to 100, leaving us with
.
(d)
; there are 51 multiples of 2, 34 multiples of 3, and 6 multiples of 17, between 1 and 102. Among these, we double-count 17 multiples of 2*3=6, 3 multiples of 2*17=34, and 2 multiples of 3*17=51; we also triple-count 1 number, 2*3*17=102. There are then 51 + 34 + 6 - (17 + 3 + 2) + 1 = 70 numbers between 1 and 102 that are not coprime to 102, and so
.
Answer:
<h2>The answer is 0.1493.</h2>
Step-by-step explanation:
In a standard deck there are 52 cards in total and there are 4 aces.
Two cards can be drawn from the 52 cards in
ways.
There are (52 - 4) = 48 cards rather than the aces.
From these 48 cards 2 cards can be drawn in
ways.
The probability of choosing 2 cards without aces is
.
The probability of getting at least one of the cards will be an ace is
.
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