1+V1+x- Find the value of dy ]
Answers
Answer:
Now , the probability that the hand contains exactly 6 face cards will be :-
\dfrac{^{12}C_{6}\times^{40}C_3}{^{52}C_9}
52
C
9
12
C
6
×
40
C
3
\begin{gathered}=\dfrac{\dfrac{12!}{6!6!}\times\dfrac{40!}{3!37!}}{\dfrac{52!}{9!\times43!}}\ \ [\because\ ^nC_r=\dfrac{n!}{r!(n-r)!}]\\\\=\dfrac{228}{91885}\end{gathered}
=
9!×43!
52!
6!6!
12!
×
3!37!
40!
[∵
n
C
r
=
r!(n−r)!
n!
]
=
91885
228
Hence, the probability that the hand contains exactly 6 face cards. is \dfrac{228}{91885}
91885
228
.
Answer:
We have,
y=
1+x
1−x
On differentiating both sides w.r.t x, we get
dx
dy
=
2
1+x
1−x
1
×(
(1+x)
2
(1+x)(0−1)−(1−x)(0+1)
)
dx
dy
=
2
1
1−x
1+x
×(
(1+x)
2
−(1+x)−(1−x)
)
dx
dy
=
2
1
1−x
1+x
×(
(1+x)
2
−1−x−1+x
)
dx
dy
=
2
1
1−x
1+x
×(
(1+x)
2
−2
)
dx
dy
=−
y
1
×
(1+x)
2
1
dx
dy
=−
y(1+x)
2
1
Hence, this is the answer.