I think you got that wrong bud, the answer should be 15 degrees.
Yeah, you right man. 15 is correct.
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WHY 15? I think it should be 30..
15 degrees because at 5:30 the hour arrow is not at 5 but in the middle between the 5 and 6
15 degrees plus or minus 180 degrees. (who knows how accurate the clock is)
45/3=15 15 is the correct answer
it is 30 degree WHY 15 ?
minTheta = 30*(360/60) = 180 hourTheta = 5*(360/12) + (360/12) * (30/60) = 165 so the difference is 15 degree
Assume a circle is 360° and you 60 grades reprsenting each one 1 minute. 360/6=6°, so 1 degree equals 1 minute At 5.30pm the arrow will be pointing at the third grade between 5 and 6 which means there are 2 left: 2*6=12° is the accurate answer ;)
No, at 5:30 the arrow will be pointing in the middle between the second and third grade = therefore its 2.5*6 = 15... or 30/2 = 15 !
the correct answer is 15 degree right?
I don't get your complicated formula... a full circle is 360 degree so a quarter of it, equal to 90 degrees. From 3-6 separated to three equal correspondent angles by 4 and 5 so 90/3 = 30 degree (clockwise) then 360 - 30 =330 degree (anticlockwise)
use normal dist to estimate binary dist, but notice the tail
binomial distribution (100 60) (1/2)^60 (1/2)^40
100 C 60 (0.5)^100 will get one prob of 60 heads, not 60 and above. Interview candidate is correct. You will most likely have to use normal dist. to estimate the binom dist given time constraints. A google search can inform one how to do this
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simply use CLT
don't have latex installed, so sorry in advance. answer is: sum[i=60 to i=100] of (100 choose i)/(2^100)
As pointed out, use a normal approximation for the sum. Define an RV called X. X = 0 for tails. X = 1 for heads. We have: E[X] = 1/2, Var[X] = 1/4. Now denote S = sum(X), where the sum runs over 100 realizations. So: E[S] = 50, Var[S] = 25. (The event are uncorrelated, so their covariance is zero, hence we can just sum variances here.) Then P(S >= 60) corresponds to P ( (S - 50)/sqrt(25) >= (60 - 50) / sqrt(25)). Note that we just subtracted the mean and divided by the standard deviation. This renders the term at the LHS a standard normal variable. So we need: P(N >= 2), where N is standard normal. If no computer is allowed, one can remember that the 97.5% percentile lies at 1.96. (If you often compute two-sided 95% confidence intervals, then you might know this number.) Or remember that within 2 sigma of the mean, 95 percent of the realizations lie. Since the distribution is symmetric, you will come to the same conclusion. So we estimate: P(S > 60) = 2.5%. In summary: make two approximations: one for going from Binomial to Normal, and a second for estimating the probability.
Using the normal approximation to a binomial distribution... Find the z value (i.e. z=x-u/st) u=mean u=np=100x0.5=50 st=standard deviation st=sq(npq)=sq(100x0.5x0.5)=5 x= 60-0.5=59.5 (i.e. continuity correction) z=59.5-50/5=1.9 Therefore P(x≥ 60)=P(z≥ 1.9)= 1-P(z≤1.9)= 1- 0.9713=0.0287 Answer: 2.87 % chance of getting more than 60 heads
"You have 25 horses, no stop watch, and you can only race 5 horses at a time. What is the minimum number of races you need in order to find the fastest horse? How would you find the second fastest horse?"
Hold 5 races, each of 5 new horses, then have a final 6th race which races all the winners of the previous 5 races. Answer: 6 races To find the second fastest, the process is the same as before except race the 2nd fastest horses from each race, the fastest horse of all the 2nd place horses is the overall second fastest.
Surely there is more to determining the second fastest horse overall... Having found the fastest horse using the method described above, after the 6th race, one would be left the fastest horse, the 4 losing horses from the 6th race and 20 other horses defeated in the first round of races. The second fastest horse is not the horse that came second in the 6th race, because there is a chance that the second fastest horse overall could have been in the same race as the eventual overall winner in the first round of races (thereby not progressing to the 6th race despite being faster than the 4 of the horses in the 6th race). Answer: So to be sure of identifying the second fastest horse overall you would have an 7th race of all the second placed horses from the first 5 races. Then, you would hold an 8th race in which the winner of the 7th race competes against the 4 losing horses from the 6th race. The winner of this race would be the second fastest overall.
Have 6 races as above for the fastest horse. Say the fastest horse overall initially won the 3rd race, then to find the 2nd fastest horse, you race the 2nd place horse from the 3rd race with the four losing horses (i.e. the four winning horses in races 1, 2, 4 & 5) from the 6th race. The winner of this 7th race is the 2nd fastest horse
A tough problem to break down. I reasoned the following way, but ended up messing up the numbers: 1) Estimate users of Dropbox in country 2) Estimate Samsung phone buyers in country 3) Estimate how many of those phone buyers are dropbox users already 4) Estimate how many of those owning the new phones, who are not dropbox users will end up activating