For the first example given the answer would be No. you have a 4 mile head start. at the 4 mile line, your friend starts at 8 miles an hour and you start running at 6 miles an hour. basic addition should give the answer. In one hour time, you wouldve ran 6 miles plus the 4 you had as a head start, giving you the 10 miles you needed to reach the finish line. He on the other hand, Biked 8 miles in an hour time. By that time, you had just reached the finish line.
So the answer is no for the first example
For the second example the maths get bit harder. You start at the 5 mile point and you friend starts at the beginning point. You only need 5 miles to win, and your friend needs double (its actually more than double, because if it was perfectly doubled, you would tie the race. Your pace just has to be a bit more than half of his speed. his speed is 17mph. yours, by logic, needs to be even a tad bit more than 8.5mph. You need to have a faster speed than 8.5mph (8.51mph works perfectly) and you win by a hair. But when we se your example, you're only going at 7mph. A whole mile and a half behind pace. Sadly, he passes you short before winning.
The second example is YES he does pass you before the end of the trail.
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Answer:
1,712,304 ways
Step-by-step explanation:
This problem bothers on combination
Since we are to select 5 subjects from a pool of 48 subjects, the number of ways this can be done is expressed as;
48C5 = 48!/(48-5)!5!
48C5 = 48!/43!5!
48C5 = 48×47×46×45×44×43!/43!5!
48C5 = 48×47×46×45×44/5!
48C5 = 205,476,480/120
48C5 = 1,712,304
Hence this can be done in 1,712,304ways
The answer is 0.6 . 3 divide 5 is 0.6
Answer:
The answer is 13 because you are a trying to put the letters alone and when you do that you eventually get 13
42 because x is equal to its corresponding side