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slava [35]
3 years ago
5

Five balls are numbered 1 through 5 and placed in a bowl. josh will randomly choose a ball from the bowl, look at its number and

then put it back into the bowl. then josh will again randomly choose a ball from the bowl and look at its number. what is the probability that the product of the two numbers will be even and greater than 10? express your answer as a common fractio
Mathematics
1 answer:
svetlana [45]3 years ago
3 0
1/5

Each of the 5 balls has 5 possible combinations, as each one (1 - 5) can be paired with any of the 5. Therefore, the total number of combinations is 25.
For the product to be even, one of the balls has to be a 2 or 4. However, 2 multiplied by any number 1 - 5 is not greater than 10, so 2 is not an option. This means one of the balls has to be a 4.
There are only 3 numbers of 1 - 5 that, when multiplied by 4, are greater than 10. These are 3, 4, and 5.
This means that there are 5 combinations that work:
3 × 4 = 12
4 × 3 = 12
4 × 4 = 16
4 × 5 = 20
5 × 4 = 20

This means that 5 out of 25 combinations produce a result that is both even and greater than 10.
5/25 is then simplified to 1/5
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dalvyx [7]

Function transformation involves changing the form of a function

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\mathbf{f(x) = 3^x}

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An exponential function is represented as:

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At point (-2,2), we have:

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At point (-1,4), we have:

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Divide both equations

\mathbf{\frac 42=\frac{ab^{-1}}{ab^{-2}}}

Simplify

\mathbf{2=\frac{b^{-1}}{b^{-2}}}

Apply law of indices

\mathbf{2=b^{-1+2}}

\mathbf{2=b}

Rewrite as:

\mathbf{b =2}

Substitute 2 for b in \mathbf{2 = ab^{-2}}

\mathbf{2 =a(2^{-2})}

This gives

\mathbf{2 =a(\frac 14)}

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Substitute 8 for (a) and 2 for (b) in \mathbf{y = ab^x}

\mathbf{y = 8(2)^x}

Express as a function

\mathbf{g(x) = 8(2)^x}

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Read more about exponential functions at:

brainly.com/question/11487261

8 0
3 years ago
What are all the exact solutions of -3tan^2(x)+1=0? Give your answer in radians.
goldenfox [79]
   
\displaystyle\\
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4 0
3 years ago
Read 2 more answers
Find the prime factorization of 2001
marusya05 [52]
Rime factorization of 2001: 

By prime factorization of 2001 we follow 5 simple steps: 
1. We write number 2001 above a 2-column table 
2. We divide 2001 by the smallest possible prime factor 
3. We write down on the left side of the table the prime factor and next number to factorize on the ride side
4. We continue to factor in this fashion (we deal with odd numbers by trying small prime factors)
5. We continue until we reach 1 on the ride side of the table

<span>2001<span>prime factorsnumber to factorize</span><span>3667</span><span>2329</span><span>291</span></span>

<span>Prime factorization of 2001 = 1×3×23×29= </span><span>1 × 3 × 23 × 29</span>
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3 years ago
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ValentinkaMS [17]

Hello There!

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4 years ago
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Answer:7.51724137931

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