@ScareCrow28@soumitrabengeri@ayushnasaif I can deduce wellacc to naga sir... since the prob to hit the target is 1/2so if we hit 6 target. obviously 3 will hit the target. and 1 more is taken to ensure that the prob is achieved. otherwise it wud hv been 1.but still with this method. prob remains less than .9please suggest frnds.
if 3 will hit( and this itself is wrong...we cant assume half will hit when the number of outcomes is so low...) Then we have a 100% prob of it destroying the target in 6 hits flat..I gave 6 as the answer on the same lines, but was pretty sure it was wrong..
In the meantime:We have 3 pastures with grass of identical height, density and growth rate. The first is 3 1/3 acres and can feed 12 oxen for 4 weeks. the second is 10 acres and can feed 21 oxen for 9 weeks. The third is 24 acres. How many oxen can be fed on these pastures for 18 weeks?
In the meantime:We have 3 pastures with grass of identical height, density and growth rate. The first is 3 1/3 acres and can feed 12 oxen for 4 weeks. the second is 10 acres and can feed 21 oxen for 9 weeks. The third is 24 acres. How many oxen can be fed on these pastures for 18 weeks?
for 6: 0.9^3* [ 2c2 + 3c2(0.1) + 4c2(0.1^2) + 5c2(0.1^3) + 6c2(0.1^4) ] =0.729 * [1+0.3+0.06+0.01+0.0015] = 0.9998235 @ScareCrow28 i thought in this way.....but anyway this is also wrng... :\
@krum bhai, ek baar isko dekho: Probability that I hit a target is 1/2. The target is destroyed if 3 shots hit the target. What is the minimum number of shots that I should try, so that the probability that I destroy the target is more than 0.9 ?1)6 2)7 3)8 4)9
Let the initial amount of grass be x and growth rate be y For the 1st pasture 10/3(x+4y)=12*4 For second 10(x+9y)=21*9 From the above equations we get x=10.8,y=0.9 For the 3rd pasture 24(10.8+18*0.9)/18=No of oxen
@ScareCrow28 bhai approach batao... i m getting 18.2 variable equation ka use kara hai kya?
We have to take into consideration the fact that the grass is growing at a constant rate.. Let k be the growth rate of the pastures. The amount of grass that grows depends on the time and the size of the pasture, grass = k*area*time --- grass = k * (10/3) * 4
but we have an initial amount of grass
So---grass = k * (10/3) * 4 + l(10/3)---(1)
Similarly for other as well... 2 variables 2 equations..you will gert the answer as 36 oxens :D
Kya baat hai!
