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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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According to the question,

x+3=3(x-5)\\\Rightarrow\ x+3=3x-15\\\Rightarrow\ 3x-x=3+15\\\Rightarrow\ 2x=18\\\Rightarrow\ x=\frac{18}{2}\\\Rightarrow\ x=9

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[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>​
asambeis [7]

Given:

\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)

To prove:

The given statement.

Proof:

We have,

\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)

LHS=\cos^4 \alpha+\sin^4\alpha

LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2

LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha     [\because a^2+b^2=(a+b)^2-2ab]

LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha      [\because \cos^2 \alpha+\sin^2\alpha=1]

LHS=1-2\cos^2 \alpha+2\cos^4 \alpha

Now,

RHS=\dfrac{1}{4}(3+\cos 4 \alpha)

RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]        [\because \cos 2\theta=2\cos^2\theta -1]

RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]

RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]        [\because \cos 2\theta=2\cos^2\theta -1]

RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]        [\because (a-b)^2=a^2-2ab+b^2]

RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]

RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]

RHS=1+2\cos^4 \alpha-2\cos \alpha

RHS=1-2\cos^2 \alpha+2\cos^4 \alpha

LHS=RHS

Hence proved.

8 0
3 years ago
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