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Elenna [48]
3 years ago
5

Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega

tive 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared?
Mathematics
1 answer:
ValentinkaMS [17]3 years ago
4 0

Answer:

\dfrac{1}{6561}

Step-by-step explanation:

Given the expression [(2^{-2})(3^{4})]^{-3} * [(2^{-3})(3^2)]^{2}, Using the laws of indices to simplify the expression. The following laws will be applicable;

a^m*a^n = a^{m+n}\\(a^m)^n = a^{mn}\\

a^{-m} = 1/a^m

Given [(2^{-2})(3^{4})]^{-3} * [(2^{-3})(3^2)]^{2}

open the parenthesis

= (2^{-2})^{-3}(3^{4})^{-3}* (2^{-3})^2(3^2)^2\\\\= 2^{-2*-3}* 3^{4*-3} * 2^{-3*2} * 3^{2*2}\\\\= 2^6 * 3^{-12} * 2^{-6} * 3^4\\\\collecting \ like \ terms\\\\= 2^6 * 2^{-6} * 3^{-12} * 3^4\\\\= 2^{6-6} * 3^{-12+4}\\\\= 2^0 * 3^{-8}\\\\= 1 * \frac{1}{3^8}\\ \\= \frac{1}{6561}

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A local technical school has 856 students. They expect 700 guests for a special speaker. The custodian has set up 1,500 chairs.
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Step-by-step explanation:

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After division we get the value  4.29310344828

Step-by-step explanation:

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A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with
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Answer:

(a) The PMF of <em>X</em> is: P(X=k)=(1-0.20)^{k-1}0.20;\  k=0, 1, 2, 3....

(b) The probability that a player defeats at least two opponents in a game is 0.64.

(c) The expected number of opponents contested in a game is 5.

(d) The probability that a player contests four or more opponents in a game is 0.512.

(e) The expected number of game plays until a player contests four or more opponents is 2.

Step-by-step explanation:

Let <em>X</em> = number of games played.

It is provided that the player continues to contest opponents until defeated.

(a)

The random variable <em>X</em> follows a Geometric distribution.

The probability mass function of <em>X</em> is:

P(X=k)=(1-p)^{k-1}p;\ p>0, k=0, 1, 2, 3....

It is provided that the player has a probability of 0.80 to defeat each opponent. This implies that there is 0.20 probability that the player will be defeated by each opponent.

Then the PMF of <em>X</em> is:

P(X=k)=(1-0.20)^{k-1}0.20;\  k=0, 1, 2, 3....

(b)

Compute the probability that a player defeats at least two opponents in a game as follows:

P (X ≥ 2) = 1 - P (X ≤ 2)

              = 1 - P (X = 1) - P (X = 2)

              =1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20\\=1-0.20-0.16\\=0.64

Thus, the probability that a player defeats at least two opponents in a game is 0.64.

(c)

The expected value of a Geometric distribution is given by,

E(X)=\frac{1}{p}

Compute the expected number of opponents contested in a game as follows:

E(X)=\frac{1}{p}=\frac{1}{0.20}=5

Thus, the expected number of opponents contested in a game is 5.

(d)

Compute the probability that a player contests four or more opponents in a game as follows:

P (X ≥ 4) = 1 - P (X ≤ 3)

              = 1 - P (X = 1) - P (X = 2) - P (X = 3)

              =1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20-(1-0.20)^{3-1}0.20\\=1-0.20-0.16-0.128\\=0.512

Thus, the probability that a player contests four or more opponents in a game is 0.512.

(e)

Compute the expected number of game plays until a player contests four or more opponents as follows:

E(X\geq 4)=\frac{1}{P(X\geq 4)}=\frac{1}{0.512}=1.953125\approx 2

Thus, the expected number of game plays until a player contests four or more opponents is 2.

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Answer:

7

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OK, so we're looking for the index in the fourth column and the fourth row

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