Question is: how many 84s will fit in 5376? Let's think about some easy multiples:
84 * 100 = 8400, so it's too big
84 * 10 = 840, so it might work
84 | 5376 | 10
-840
84 | 4536 | 10
-840
84 | 3696 | 10
-840
84 | 2856 | 10
-840
84 | 2016 | 10
-840
84 | 1176 | 10
-840
84 | 336
We can't fit any more 840 in 336, so we check how many 84s are in 336 and what's the remainder:
84 | 336 | 4
- 336
So there's no remainder. Now we add all the partial quotients to get the final result:
10 + 10 + 10 + 10 + 10 + 10 + 4 = <u>64
</u>It's correct, I checked it with calculator. I just hope you'll be able to read something from that, it's quite difficult to do partial dividing with no pencil and paper :)
Answer:
(E) 0.83
Step-by-step explanation:
We will solve it using conditional probability.
Let A be the event that a TV show is successful.
P(A) = 0.5
A' be event that the show is unsuccessful
P(A') =0.5
Let B be the event that the response was favorable
P(B) = 0.6
Let B' be the event that the response was unfavorable/
P(B') = 0.4
P(A∩B) = 0.5 and P(A∩B') = 0.3
We need to find new show will be successful if it receives a favorable response.
P(A/B) = 
= 0.5/0.6
= 0.833
A square with sides of 4 for perimeter and area of a rectangle with length 2 width 6<span />
Answer:

Step-by-step explanation:
GIVEN: two two-letter passwords can be formed from the letters A, B, C, D, E, F, G and H.
TO FIND: How many different two two-letter passwords can be formed if no repetition of letters is allowed.
SOLUTION:
Total number of different letters 
for two two-letter passwords
different are required.
Number of ways of selecting
different letters from
letters


Hence there are
different two-letter passwords can be formed using
letters.
Answer:
-6
Step-by-step explanation:
3x-3(x+1)-3
3x-3x-3-3
-3-3
-6