Question 1
Answer
f (x) = 𝑥 2 − 𝑥
The slope of the tangent line is given by f’(x).
𝑑[𝑥 2 −𝑥]
f’(X) =
𝑑𝑥
𝑑𝑥 2 𝑑𝑥
= -
𝑑𝑥 𝑑𝑥
=2x-1
The slope of the tangent line at the point x= 2 is given by f’ (2).
Slope= f’(2)
=2×2-1
=3
The correct option is given by
o b.3
,Question 2
Answer
Given,
E(X) = 6
Np = 6
Var(X) = 4.2
Npq = 4.2
So,
𝑛𝑝 6
=
𝑛𝑝𝑞 4.2
We get,
q= 0,7
p=1-q=0.3
Therefore,
P=0.3
And
, 6
N= =20
0.3
So, n=20 and p=0.3
P(X ≤ 6) = P( X = 1) + P( X = 2 ) + P( X = 3 ) + P( X = 4 ) + P( X = 5 ) + P( X
=6)
We use the formula
𝑛
P(X= r) = ( ) 𝑝𝑟 𝑞 𝑛−𝑟
𝑟
Therefore,
0.0798
P(X=0) =
100
0.684
P(X=1) =
100
2.7846
P(X=2) =
100
7.16
P(X=3) =
100
13.042
P(X=4) =
100
17.886
P(X=5) =
100
19.164
P(X=6) =
100
60.8004
P(X≤ 6) =
100
P(X≤ 6) = 0.608
ANS -> TRUE(T)
Answer
f (x) = 𝑥 2 − 𝑥
The slope of the tangent line is given by f’(x).
𝑑[𝑥 2 −𝑥]
f’(X) =
𝑑𝑥
𝑑𝑥 2 𝑑𝑥
= -
𝑑𝑥 𝑑𝑥
=2x-1
The slope of the tangent line at the point x= 2 is given by f’ (2).
Slope= f’(2)
=2×2-1
=3
The correct option is given by
o b.3
,Question 2
Answer
Given,
E(X) = 6
Np = 6
Var(X) = 4.2
Npq = 4.2
So,
𝑛𝑝 6
=
𝑛𝑝𝑞 4.2
We get,
q= 0,7
p=1-q=0.3
Therefore,
P=0.3
And
, 6
N= =20
0.3
So, n=20 and p=0.3
P(X ≤ 6) = P( X = 1) + P( X = 2 ) + P( X = 3 ) + P( X = 4 ) + P( X = 5 ) + P( X
=6)
We use the formula
𝑛
P(X= r) = ( ) 𝑝𝑟 𝑞 𝑛−𝑟
𝑟
Therefore,
0.0798
P(X=0) =
100
0.684
P(X=1) =
100
2.7846
P(X=2) =
100
7.16
P(X=3) =
100
13.042
P(X=4) =
100
17.886
P(X=5) =
100
19.164
P(X=6) =
100
60.8004
P(X≤ 6) =
100
P(X≤ 6) = 0.608
ANS -> TRUE(T)