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New answer posted

6 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

3. Given,

P(A)=0.8

P(B)=0.5

P(B/A)=0.4

(i) P(AB)

P(B/A)=P(BA)P(A) [?P(AB)=P(BA)]

0.4=P(AB)0.8

P(AB)=0.4*0.8

=0.32

(ii) P(A/B)

P(A/B)=P(AB)P(B)

P(A/B)=0.320.5

=0.64

(iii) P(AB)

P(AB)=P(A)+P(B)P(AB)

=0.8+0.50.32

=0.98

New answer posted

6 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

2. Given,

P (B)=0.5

P (AB)=0.32

P (A/B)=P (AB)P (B)=0.320.5=3250=1645

New answer posted

6 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

1. Given,

P (E)=0.6

P (F)=0.3

P (EF)=0.2

P (E/F)=P (EF)P (F)=0.20.3=23

P (F/E)=P (FE)P (E)=0.20.6=26=13

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

36. For the given inequality, x> –3.

The equation of the line is x= –3.

This line divides the xy-plane into planer I and II. We take a point (0,0) to check the correctness of the inequality.

So, 0> –3 which is true.

So, the solution of the region is I which includes the origin.

The dotted line indicates that any point on the line does not satisfy the inequality.

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

35. For the given equation inequality y< 2,  the equation of line is y= 2

This line devides the xy-plane into two planer I and II. We take a point (0,0) to check the correctness of the inequality.

So, 0< 2

0< 2 which is false.

So, the solution of the region is II which does not include the origin.

The dotted line indicates that any point on the line does not satisfy the inequality.

New answer posted

6 months ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

34. For the inequality 3y – 5x<30, the equation of line is 3y 5x=30. We consider the table below to plot 3y 5x =30.

xy|010|60|

This line divides the xy-plane into two planer I and II. We select a point (0,0) and check the correctness the inequality.

3 * 0 – 5 * 0 < 30

0 < 30 which is true.

So, the solution region is I which includes the origin. The dotted line indicates that any point on the line will not satisfy the given inequality.

New question posted

6 months ago

0 Follower 2 Views

New answer posted

6 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

33.For the inequality –3x+2y≥ –6 the equation of line is – 3x+2y=6.

We consider the table below to plot – 3x+2y= –6.

xy|03|20|

This line divides the xy-plane into two planer I and II. We select a point (0,0) and check the correctness the inequality,

–3 * 0+2 * 0 ≥ –6

0 ≥ –6 which is true.

So, the solution region is I which includes the origin. The continuous line also indicates that any point on the line also satisfy the given inequality.

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

32. For the inequality, 2x – 3y>6 the eqn is 2x – 3y=6. We consider the table below to plot 2x – 3y=6.

xy|02|30|

This line divides the xy-plane into half planer I and II. We select point (0,0) and check the correctness of the inequality.

2 * 0 – 3 * 0>6

0>6 which is false.

So, the solution region is II which does not includes the origin (0,0).

The dotted line indicates that the any point on the line does not satisfy the given inequality.

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

31. For the inequality x – y≤ 2 so the equation of line is x – y=2. We consider the table below to plat x – y=2

xy|02|20|

This line divides the xy-plane into half planer I and II. We select point (0,0) and check the correctness of the inequality.

0 – 0 ≤ 2

0 ≤ 2 which is true.

So, the solution region is I which includes the origin (0,0). The continuous line also indicates that any point on the line also satisfy the given inequality.

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