- Messages
- 1,171
- Reaction score
- 4,151
- Points
- 273
Well you can also do it by plotting a speed time graph and then calculating the area under graphthen distance is speed x time so 604.9 x 15 = 9073.5m
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
Well you can also do it by plotting a speed time graph and then calculating the area under graphthen distance is speed x time so 604.9 x 15 = 9073.5m
I'm sorry I forgot to tell you that you can't use distance = speed/time formula here because you aren't given an average speed.then distance is speed x time so 604.9 x 15 = 9073.5m
(x^2 + y^2) = 2(x^2 – y^2)
Co-ordinates of M ( maximum point)
Solve the square. And the parenthesis.
x^4 +2(x^2)(y^2) + y^4 = 2(x^2) – (2y^2)
Differentiate.
Dy/dx ,
4(x^3) + 2[2*x*(y^2) +2*y*(x^2) (dy/dx)] +4(y^3)(dy/dx) = 4x – 4y(dy/dx)
4(x^3) + 4*x*(y^2) + 4*y*(x^2)(dy/dx) +4(y^3)(dy/dx) = 4x – 4y(dy/dx)
Make dy/dx the subject,
Dy/dx[4*y*(x^2) +4*(y^3) - 4y]= 4x – 4*x*(y^2) -4(x^3)
Dy/dx = (4x – 4*x*(y^2) -4*(x^3))/( 4*y*(x^2) +4*(y^3) - 4y)
Equate to zero since to find stationary points we put dy/dx=0
0 = (4x – 4*x*(y^2) - 4*(x^3))/( 4*y*(x^2) +4*(y^3) - 4y)
All 4’s will cancel.
0 = (x – x*(y^2) - (x^3))
0 = x(1 – (y^2) - (x^2))
0 = 1 – (y^2) – (x^2)
-1 = - (y^2) – (x^2)
( x^2) + (y^2) = 1 & (x^2) = 1 – (y^2)
Now use the equation of the curve and plug in he above value.
((x^2) + ( y^2)) = 2((x^2) – ( y^2))
(1)^2 = 2(1 - (y^2) - (y^2))
1 = 2(1 – 2(y^2))
½ = 1 – 2(y^2)
½ - 1 = - 2(y^2)
- 1/2 = - 2y^2
(1/2)/2 = y^2
y^2 = ¼
y = sqrt(1/4)
y = ½
x ^2 = 1 – y^2
x ^2 = 1 – (1/2)^2
x ^2 = 1 - ¼
x ^2 = ¾
x= sqrt(3/4)
x= (sqrt3)/2 Ans.
Whats the answer? Ill post my solution if I got it rightCan anyone please help me solve this:
A choir has seven sopranos, six altos, three tenors and four bases. The sopranos and altos are women and the tenors and basses are men. At a particular rehearsal, three members of the choir are chosen at random to make the tea.
(a) Find the probability that all three tenors are chosen.
(b) Find the probability that exactly one bass is chosen.
(c) Find the conditional probability that two women are chosen, given that exactly one bass is chosen.
(d) Find the probability that the chosen group contains exactly one tenor or exactly one bass (or both).
Ahhh okay, thank you soo muchI'm sorry I forgot to tell you that you can't use distance = speed/time formula here because you aren't given an average speed.
You SHOULD plot a speed time graph and calculate the area under it to get the right answer. See the pic plz
Okay I triedCan anyone please help me solve this:
A choir has seven sopranos, six altos, three tenors and four bases. The sopranos and altos are women and the tenors and basses are men. At a particular rehearsal, three members of the choir are chosen at random to make the tea.
(a) Find the probability that all three tenors are chosen.
(b) Find the probability that exactly one bass is chosen.
(c) Find the conditional probability that two women are chosen, given that exactly one bass is chosen.
(d) Find the probability that the chosen group contains exactly one tenor or exactly one bass (or both).
Thanks !Okay I tried
Note: 3 are chosen from a total of 20 so your total is 20C3 = 1140(useful to find probability)
a) 3C3*17C0 = 1
For probability, divide by total
1/1140 = 0.00087
b) List all the possibilities. Remember 3 are chosen and exactly 1 bass is chosen….
● 4C1*3C2 = 12
● 4C1*6C2 = 60
● 4C1*7C2 = 84
●4C1*3C1*6C1 = 72
●4C1*3C1*7C1 = 84
●4C1*7C1*6C1 = 168
Add all and divide by total = 0.421 or 8/19
Shortcut: (4C1*16C2)/20C3 = 0.421 or 8/19
(1 from 4 bases and 2 from the remaining 16)
c) 2 women = 13C2*4C1 + 13C2*3C1 = 546
546/1140 = 91/190
P(2 women)/p(exactly 1 base(ans to b)) =(91/190)/(8/19) (This is probably wrong, idk)
d) Incase of ‘or’ use intersection
Neither base nor tenor = 3C1*17C2 + 4C1*16C2 - 12*13 = 732
732/1140 = 0.642 Ans.
Okay I tried
Note: 3 are chosen from a total of 20 so your total is 20C3 = 1140(useful to find probability)
a) 3C3*17C0 = 1
For probability, divide by total
1/1140 = 0.00087
b) List all the possibilities. Remember 3 are chosen and exactly 1 bass is chosen….
● 4C1*3C2 = 12
● 4C1*6C2 = 60
● 4C1*7C2 = 84
●4C1*3C1*6C1 = 72
●4C1*3C1*7C1 = 84
●4C1*7C1*6C1 = 168
Add all and divide by total = 0.421 or 8/19
Shortcut: (4C1*16C2)/20C3 = 0.421 or 8/19
(1 from 4 bases and 2 from the remaining 16)
c) 2 women = 13C2*4C1 + 13C2*3C1 = 546
546/1140 = 91/190
P(2 women)/p(exactly 1 base(ans to b)) =(91/190)/(8/19) (This is probably wrong, idk)
d) Incase of ‘or’ use intersection
Neither base nor tenor = 3C1*17C2 + 4C1*16C2 - 12*13 = 732
732/1140 = 0.642 Ans.
Sorry for the late reply. My internet loses it sometimes. No wonder when it might lose it nextI'd solved the (c) part. Here goes its solution:
P(2 women are chosen | exactly one bass) = ( 6 * (13/20 * 12/19 * 4/18) ) / 0.421 = 0.650
But I was having a problem in the last part. Your answer is correct! But can you please tell me what's the problem with my working? I did this way:
( 4C1 * 13C2 + 3C1 * 13C2 + 3C1 * 4C1 * 13C1 ) / 20C3 = 0.616
And please do explain your working for last part in a bit more detail.
v1(t) = 16*t - 6*t^2two objects move along the same straight line. the velocities of the objects are given by v1=16t - 6t^2 and v2=2t-10. initially the objects are 32m apart. at what time do they collide? the answer is 2s... could anyone explain this to me?
integrate both eq.s and u ll get their distances.two objects move along the same straight line. the velocities of the objects are given by v1=16t - 6t^2 and v2=2t-10. initially the objects are 32m apart. at what time do they collide? the answer is 2s... could anyone explain this to me?
Thank you!!!v1(t) = 16*t - 6*t^2
s1(t) = -2*t^2*(t - 4)
v2(t)= 2*t-10
s2(t) = t*(t - 10)
So we know the displacement function for both objects. They are 32m apart initially, so one must travel 32m more than the other to meet each other:
s1 = s2 + 32
ans =
2.000000000000000
3.676174977679906
-2.176174977679906
Thank you!!integrate both eq.s and u ll get their distances.
s1 = 8t^2 - 2t^3 + c
s2 = t^2 - 10t + k
since u dont have any origin, u assume a point as origin.
lets take initial position ob object 1 for that here.
at time t = 0,
s1 = 0 = 8(0)^2 - 2(0)^3 + c
hence c = 0
(w.r.t. origin i.e. s1) s2 = 32 = (0)^2 - 10(0) + k
hence k = 32
eq we get are
s1 = 8t^2 - 2t^3
s2 = t^2 - 10t + 32
put s2- s1 = 0, simplify and u ll get...
2t^3 - 7t^2 - 10t + 32 = 0
solve the eq (with hit and trial of course) and u'll have t = 2
Thank you so much! I got it! Actually I was totally ignoring some of the possible arrangements; my bad Anyway thanks once again.Sorry for the late reply. My internet loses it sometimes. No wonder when it might lose it next
So. You have
Ways of choosing exactly one tenor. ANY of the remaining for the other two people
3C1*17C2 = 408
You did 3C1 * 13C2. Why 13? total people are 20. You have to choose 1 from 3 and the other 2 from remaining 17.
Similarly, ways of choosing exactly one bass.
4C1*16C2= 480
Now the question says '(or both)'
Ways of choosing both
3C1*4C1*13C1 = 156
Now look. If you have 'or' in the question you will look for intersection you will get this by adding 408 and 480 and then subtracting 156 from the answer. If you wont, you will be double counting. If the question said 'and' then you would go for addition (union) Therefore
408 + 480 - 156 = 732
Now for probability divide by total.
732/1140= 0.642
I hope I'm coherent enough. Should you require any more detail, feel free to ask!
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now