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wolverine [178]
3 years ago
5

Factorise 6x^2 - 21x

Mathematics
1 answer:
astraxan [27]3 years ago
7 0
<span>6x^2 - 21x = 3x(2x - 7)

hope it helps</span>
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Answer:

  64/49

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)^c = a^(bc)

  (a^b)(a^c) = a^(b+c)

  a^-b = 1/a^b

  a^0 = 1

_____

Using these rules, we can simplify the given expression to ...

  \left(\dfrac{8\cdot 4\cdot 2}{8\cdot 7}\right)^2\times\left(\dfrac{8^0}{7^{-3}}\right)^3\times 7^{-9}\\\\=\dfrac{8^2}{7^2}\times(8^{0\cdot 3}\cdot 7^{3\cdot 3})\times 7^{-9}\\\\=\dfrac{64}{49}\times(8^07^{9-9})=\boxed{\dfrac{64}{49}}

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3 years ago
A box contains 20 light box of which five or defective it for lightbulbs or pick from the box randomly what's the probability th
Snowcat [4.5K]

Answer:

1

Step-by-step explanation:

Given:-

- The box has n = 20 light-bulbs

- The number of defective bulbs, d = 5

Find:-

what's the probability that at most two of them are defective

Solution:-

- We will pick 2 bulbs randomly from the box. We need to find the probability that at-most 2 bulbs are defective.

- We will define random variable X : The number of defective bulbs picked.

Such that,               P ( X ≤ 2 ) is required!

- We are to make a choice " selection " of no defective light bulb is picked from the 2 bulbs pulled out of the box.

- The number of ways we choose 2 bulbs such that none of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 2:

        X = 0 ,       Number of choices = 15 C r = 15C2 = 105 ways

- The probability of selecting 2 non-defective bulbs:

      P ( X = 0 ) = number of choices with no defective / Total choices

                       = 105 / 20C2 = 105 / 190

                       = 0.5526

- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:

        X = 1 ,       Number of choices = 15 C 1 * 5 C 1 = 15*5 = 75 ways

- The probability of selecting 1 defective bulbs:

      P ( X = 1 ) = number of choices with 1 defective / Total choices

                       = 75 / 20C2 = 75 / 190

                       = 0.3947

- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.

        X = 2 ,       Number of choices = 5 C 2 = 10 ways

- The probability of selecting 2 defective bulbs:

      P ( X = 2 ) = number of choices with 2 defective / Total choices

                       = 10 / 20C2 = 10 / 190

                       = 0.05263

- Hence,

    P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)

                     = 0.5526 + 0.3947 + 0.05263

                     = 1

7 0
3 years ago
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