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il63 [147K]
2 years ago
9

A bag contain 3 white balls, 4 blue balls and 2 red balls. Two balls are to be taken at random from the bag. The first ball is n

ot replaced before the secod one is taken. Find the probability that
a) the first ball taken will be blue

b) both balls taken will be blue

c) both balls taken will be the same colour
Mathematics
1 answer:
Liono4ka [1.6K]2 years ago
3 0

Answer:

a)the first ball taken will be blue

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I NEED HELP PLEASE, THANKS! :)
Afina-wow [57]

Answer:

Third option is the right choice.

Step-by-step explanation:

Manipulate the right hand side.

\sec \left(x\right)\\\\=\frac{1}{\cos \left(x\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \:1=\cos ^2\left(x\right)+\sin ^2\left(x\right)\\\\=\frac{\cos ^2\left(x\right)+\sin ^2\left(x\right)}{\cos \left(x\right)}\\\\\Rightarrow \mathrm{True}

Best Regards!

6 0
3 years ago
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A chef cooks rice using a 1: 2 ratio of dry rice to water. Her recipe calls for 9.5 cups of dry rice.
tigry1 [53]

Answer:

19 cups of water.

Step-by-step explanation:

1:2 ratio

1=9.5

2=9.5*2= 19

8 0
2 years ago
Please tell me the answer and explain
k0ka [10]

Answer:

answer is

{X^2 + 4, X<2

{X + 4, X>2

Step-by-step explanation:

Because both solution set graph is at the begining hole at the value of X = 2 this means both sets doesn't contain 2.

7 0
1 year ago
Hey would this be 36 ? ..the reason I am asking is because I'm trying to help a friend and the closest answer on the bottom is 3
damaskus [11]
A=(7*3)+(15-7)*2

A=21+2*8

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8 0
3 years ago
Read 2 more answers
A particle moves with velocity vector
asambeis [7]

Answer:

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}

Step-by-step explanation:

We are given that velocity vector of a particle

\vec{v(t)}=t\hat{i}-2\hat{j}+t^2\hat{k}

When t=0 then the particle is at the point (0,-1,1).

We have to find the position of particle  at time t.

We know that

Velocity =\frac{Displacement }{time}=\frac{ds}{dt}

Therefore,\vec{v}=\frac{\vec{ds}}{dt}

\int{ds}=\int (t\hat{i}-2\hat{j}+t^2\hat{k})dt

Integrate on both sides then we get

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}+C

\int x^n dx=\frac{x^{n+1}}{n+1}+C

Substitute the value of point at time t=0 then we get

C=-\hat{j}+\hat{k}

Substitute the value of C then we get

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}-\hat{j}+\hat{k}

Therefore, the position of particle at time t

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}

8 0
2 years ago
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