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Norma-Jean [14]
2 years ago
15

In one lottery game, contestants pick five numbers from 1 through 23 and have to match all five for the big prize (in any order)

. You'll get twice your money back if you match three out of five numbers. If you buy four tickets, what's the probability of matching three out of five numbers?
(Enter your answer as a fraction in lowest terms.)
Mathematics
1 answer:
mihalych1998 [28]2 years ago
6 0

The probability of matching three out of five numbers is 153/33649

<h3>How to determine the probability?</h3>

The numbers are given as:

1 to 23

There are five matching numbers

The  probability of getting a match is:

p = 5-k/n where k = 0, 1, 2

While the complement probability is

q = (n - 2)/n

Using the above formulas, the probability of getting from 5 is:

P = 5/23 * 4/22 * 3/21 * 18/20 * 17/19

Evaluate the product

P = 153/33649

Hence, the probability of matching three out of five numbers is 153/33649

Read more about probability at:

brainly.com/question/14290572

#SPJ1

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Mrrafil [7]

Answer:

Infinite number of solutions

Step-by-step explanation:

2(3 + 4x) = 8x + 6\\6 + 8x = 8x+6\\8x-8x=6-6\\0=0

For any value of x, both equations are equal to each other because both sides are identical

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Answer:

  1. b/a
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Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

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1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

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2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

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3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

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4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

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5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}

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Answer:

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Step-by-step explanation:

<u>Step 1: Define</u>

9b⁶c⁵

<u>Step 2 Identify</u>

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Answer:

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which will give as ;

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