Mathematics: Post your doubts here!

[The Sarcastic Retard]

i already got it, the answer for the time was 605s and the distance 9075m

Post the solution.

 

[Maayee]

When the car is accelerating:

s = (v²-u²)/2a
= (15²-0)/2(5) = 22.5 m
time taken for acceleration
t = (v-u)/a
= (15-0)/5 = 3 s

When the car is speeding uniformly:
s = velocity x time = 15 x 10 min x 60s = 9000m

When the car is decelertating:

s = (v²-u²)/2a
= (0-15²)/2(-8) = 14.1 m
time taken for decelartion
t = (v-u)/a
= (0-15)/(-8) = 1.9 s


Total journey time = 3 + (10x60) + 1.9 = 604.9s

 

[Maayee]

then distance is speed x time so 604.9 x 15 = 9073.5m

 

[Mr.Physics]

then distance is speed x time so 604.9 x 15 = 9073.5m

Well you can also do it by plotting a speed time graph and then calculating the area under graph

 

[Mr.Physics]

then 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.
You SHOULD plot a speed time graph and calculate the area under it to get the right answer. See the pic plz

 

[Rizwan Javed]

Can 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).

 

[unhellty]

(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.

Thank you!

 

[Anum96]

Can 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).

Whats the answer? Ill post my solution if I got it right

 

[Maayee]

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.
You SHOULD plot a speed time graph and calculate the area under it to get the right answer. See the pic plz

Ahhh okay, thank you soo much

 

[Anum96]

Can 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).

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.

 

[Rizwan Javed]

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.

Thanks !

 

[Rizwan Javed]

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.

I'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.

 

[unhellty]

In (ii), why can't I do it this way:

Let X be Mr Parry; Y be Mrs Parry.

after changing to miles > X~N(320,2402.5); Y~N(55.625, 34.225)

E(X+Y) = 320+556.25 = 55.625

Var(X+Y) = 2402.5+34.225 = 2436.725
hence, my ans to std deviation is 49.36.

But the working ans is:
Var[5/8(Mr + Mrs)] = (25/64)×3898.76 = 1520
the ans is 39.0 miles

What happens if I change to miles first then only sum it up? Is it because it is a multiple distribution instead of a sum? (Idk if that's a thing but that's what my school teacher said ._.)

 

[unhellty]

Actually I can't rly differentiate between the multiple and sum thingy Anybody care to explain?

 

[Maayee]

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?

 

[Anum96]

I'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.

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!

 

[qwertypoiu]

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?

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

 

[***amd***]

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?

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

 

[bakhita]

Which book should I use for Trignometry? I've been using Core Maths by L.Bostock and S. Chandler so far, it has been pretty fine until now, but I'm like at a dead end in trignometry. Any suggestions?

 

[happyoner_5]

plzz help ASAP..!!