ps questions - 2

I am not too sure ..but i think this is how it shd be ..
17 defects in 51000 chips

x in 60000

51000x = 17 * 60000
x = 20 ..

Take root80 = 9 then it will be simpler

(root 9 + 9)^2

roots cancel out hence 18

xcise, is the answer to the 2nd question B (9-4root5)????

Couldnt be D 18??

no, sorry about that. the answer is 18 (D)

Yes that is what i meant : ). I get 18 (D), but you had selected answer B.

isn't the answer E (20) ?

If you solve it as a^2 + 2ab + b^2,

9 + rt(80) + 2 *rt((9 +rt(80)) * (9-rt(80)) + 9 - rt(80)
= 18 + 2* rt(81 -80) = 18 +2 =20

am i missing something here?

Guys, sorry! I am agree with malnv! is it not 20 instead of 18? Are we missing something? sugesstion welcomed and obviously required ...

malnv wrote:isn't the answer E (20) ?

If you solve it as a^2 + 2ab + b^2,

9 + rt(80) + 2 *rt((9 +rt(80)) * (9-rt(80)) + 9 - rt(80)
= 18 + 2* rt(81 -80) = 18 +2 =20

am i missing something here?

This solution is what I got as well.

I agree with malnv the answer is 20 instead of 18

applying (a+b)^2=a^2 + 2ab + b^2 to the question we have
9 + sqrt(80) + 2 sqrt((9 +rt(80)) * sqrt(9-rt(80)) + 9 - sqrt(80)
since sqrt(x)*sqrt(y)=sqrt(xy) then we have
18 + 2 sqrt(81 -80)
=18+2sqrt(1)
=20

Suggest me if Iam Wrong.

Regards
Sibbineni

Question 1:
Let E be the expected number
k=number of tosses
p=probability of success

then the expected number
E=kp is the formula

so applying the aboe formula we have

In the question k=60,000

p= is the number of defective chips /total number of the chips in the shipment=17/51000

E=60,000*17/51000

E=20

Suggest me if Iam wrong

Regards
Sibbineni