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kotykmax [81]
2 years ago
14

Slips of paper numbered 1 to 15 are placed in a box. A slip of paper is drawn at random. What is the probability that the number

picked is either a multiple of 5 or an odd number?
Mathematics
1 answer:
monitta2 years ago
8 0

Given:

Slips of paper numbered 1 to 15 are placed in a box.

To find:

The probability that the number picked is either a multiple of 5 or an odd number.

Solution:

We have,

Total outcomes = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

No. of total outcomes = 15

Multiple of 5 are 5, 10, 15.

Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15.

Number that are either a multiple of 5 or an odd number are 1, 3, 5, 7, 9, 10, 11, 13, 15.

No. of favorable outcomes = 9

We know that,

\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}

\text{Probability}=\dfrac{9}{15}

\text{Probability}=\dfrac{3}{5}

\text{Probability}=0.6

Therefore, the  probability that the number picked is either a multiple of 5 or an odd number is 0.6.

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A number cube with the numbers 1 through 6 is rolled. What is the probability of the number cube showing a 4 or a number less th
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Answer:

Step-by-step explanation:

The probability is

P

(

A

)

=

1

3

Explanation:

To calculate the probability you have to count the number of all possible results

|

Ω

|

and the number of results that fulfill the given condition

|

A

|

.

In this case

Ω

=

{

1

,

2

,

3

,

4

,

5

,

6

}

Thereare 6 possible result of a dice toss. So

|

Ω

|

=

6

The given condition is "the result is divisible by

3

", so

A

=

{

3

,

6

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3

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|

A

|

=

2

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Finally to calculate the probability we have to divide

|

A

|

by

|

Ω

|

.

P

(

A

)

=

|

A

|

|

Ω

|

=

2

6

=

1

3

Note The probability is never larger than

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3 years ago
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Answer:

77.  \cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} xProved

78.  \sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ] Proved

79. \cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x Proved.

80. \sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x Proved.

Step-by-step explanation:

77. Left hand side

= \cot^{6} x

= \cot^{4} x \times \cot^{2} x

= \cot^{4}x [\csc^{2}x - 1]  

{Since we know, \csc^{2} x - \cot^{2}x = 1}

= \cot^{4} x \csc^{2}x - \cot^{4} x  

= Right hand side (Proved)

78. Left hand side

= \sec^{4}x \tan^{2} x

= \sec^{2} x [1 + \tan^{2}x] \tan^{2} x  

{Since \sec^{2}x - \tan^{2}x = 1}

= \sec^{2}x [\tan^{2}x + \tan^{4}x ]

= Right hand side (Proved)

79. Left hand side  

= \cos^{3} x\sin^{2} x

= \cos x[1 - \sin^{2} x] \sin^{2} x

{Since \sin^{2}x + \cos^{2} x = 1}

= [\sin^{2}x - \sin^{4}x] \cos x

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80. Left hand side  

= \sin^{4}x - \cos^{4}x

= [\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x

{Since \sin^{2}x + \cos^{2} x = 1}

= 1 - 2\cos^{2} x[1 - \cos^{2}x ]

= 1 - 2\cos^{2}x + 2 \cos^{4} x

= Right hand side. (Proved)

7 0
3 years ago
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