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  1. Asked: November 23, 2020In:

    Simplify \dfrac{4x^{2}-49y^{2}}{2x^{2}+xy-7y^{2}}

    blessing_O
    blessing_O Moderator

    $\dfrac{4x^{2}-49y^{2}}{2x^{2}+xy-7y^{2}}$ The numerator is difference of two squares $\therefore 4x^{2}-49y^{2}=(2x)^{2}-(7y)^{2}=(2x-7y)(2x+7y)$ The denominator is a quadratic equation but it can not be factorized $\therefore \dfrac{4x^{2}-49y^{2}}{2x^{2}+xy-7y^{2}}=\dfrac{(2x-7y)(2x+7y)}{2x^{2}+xRead more

    \dfrac{4x^{2}-49y^{2}}{2x^{2}+xy-7y^{2}}
    The numerator is difference of two squares
    \therefore 4x^{2}-49y^{2}=(2x)^{2}-(7y)^{2}=(2x-7y)(2x+7y)
    The denominator is a quadratic equation but it can not be factorized
    \therefore \dfrac{4x^{2}-49y^{2}}{2x^{2}+xy-7y^{2}}=\dfrac{(2x-7y)(2x+7y)}{2x^{2}+xy-7y^{2}}

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  2. Asked: December 20, 2020In:

    The probability of obtaining an odd number in a single toss of a fair die is what?

    blessing_O
    Best Answer
    blessing_O Moderator

    Sample space=$(1,2,3,4,5,6)$ odd numbers=${1,3,5}$ $\therefore$ P(odd numbers)=$\dfrac{3}{6}=\dfrac{1}{2}$

    Sample space=(1,2,3,4,5,6)
    odd numbers={1,3,5}
    \therefore P(odd numbers)=\dfrac{3}{6}=\dfrac{1}{2}

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  3. Asked: December 20, 2020In:

    What is the propability of obtaining a odd number in a single toss of a fair die?

    blessing_O
    Best Answer
    blessing_O Moderator

    Sample space=$(1,2,3,4,5,6)$ odd numbers=${1,3,5}$ $\therefore$ P(odd numbers)=$\dfrac{3}{6}=\dfrac{1}{2}$

    Sample space=(1,2,3,4,5,6)
    odd numbers={1,3,5}
    \therefore P(odd numbers)=\dfrac{3}{6}=\dfrac{1}{2}

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  4. Asked: December 20, 2020In:

    What is the propability of obtaining a prime number in a single toss of a fair die?

    blessing_O
    Best Answer
    blessing_O Moderator

    Sample space=$(1,2,3,4,5,6)$ Prime numbers=${2,3,5}$ $\therefore$ P(Prime numbers)=$\dfrac{3}{6}=\dfrac{1}{2}$

    Sample space=(1,2,3,4,5,6)
    Prime numbers={2,3,5}
    \therefore P(Prime numbers)=\dfrac{3}{6}=\dfrac{1}{2}

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  5. Asked: December 20, 2020In:

    What is the propability of obtaining a multiple of 2 in a single toss of a fair die?

    blessing_O
    Best Answer
    blessing_O Moderator

    Sample space=${1,2,3,4,5,6}$ multiples of 2=${2,4,6}$ $\therefore$ P(multiples of 2)=$\dfrac{3}{6}=\dfrac{1}{2}$

    Sample space={1,2,3,4,5,6}
    multiples of 2={2,4,6}
    \therefore P(multiples of 2)=\dfrac{3}{6}=\dfrac{1}{2}

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  6. Asked: December 20, 2020In:

    The propability that a number selected at random from 1 to 40 is a multiple of 7 is?

    blessing_O
    Best Answer
    blessing_O Moderator

    Sample space=${1,2,3,4,5,6...40}$ multiple of 7 within 1 and 40=${7,14,21,28,35}$ $\therefore$ P(multiple of 7)=$\dfrac{5}{40}=\dfrac{1}{8}$

    Sample space={1,2,3,4,5,6...40}
    multiple of 7 within 1 and 40={7,14,21,28,35}
    \therefore P(multiple of 7)=\dfrac{5}{40}=\dfrac{1}{8}

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