So i'm assuming 33^a is saying any number that is a square of 33 inside of 1,313 so first, to find how many even numbers divide 1,313 by two to get 656.5 so if I am not mistaken 656 numbers in 1,313 will be even. Now find all squares of 33 that don't pass 1,313 which is 33 itself and 1,089 because 33^3 is 35,937 so add 2 to 656 to get 668. so the probability is 658/1,313 i believe.
21 should be the right answer
Notice that
(1 + <em>x</em>)(1 + <em>y</em>) = 1 + <em>x</em> + <em>y</em> + <em>x y</em>
So we can add 1 to both sides of both equations, and we use the property above to get
<em>a</em> + <em>b</em> + <em>a b</em> = 76 ==> (1 + <em>a</em>)(1 + <em>b</em>) = 77
and
<em>c</em> + <em>d</em> + <em>c d</em> = 54 ==> (1 + <em>c</em>)(1 + <em>d</em>) = 55
Now, 77 = 7*11 and 55 = 5*11, so we get
<em>a</em> + 1 = 7 ==> <em>a</em> = 6
<em>b</em> + 1 = 11 ==> <em>b</em> = 10
(or the other way around, since the given relations are symmetric)
and
<em>c</em> + 1 = 5 ==> <em>c</em> = 4
<em>d</em> + 1 = 11 ==> <em>d</em> = 10
Now substitute these values into the desired quantity:
(<em>a</em> + <em>b</em> + <em>c</em> + <em>d</em>) <em>a</em> <em>b</em> <em>c</em> <em>d</em> = 72,000
Either 1/10 or 10%
Hope this helps!!!
Answer:
5 to 3
Step-by-step explanation:
