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A flower vase has 5 white lilies, 4 pink roses, and 6 yellow carnations.

A flower vase has 5 white lilies, 4 pink roses, and 6 yellow carnations. One flower is chosen at random and given to a woman. Another flower is then chosen at random and given to a different woman. What is the probability that both flowers are pink roses?

2 months ago

Solution 1

Guest Guest #3671
2 months ago
Total flowers = 5 + 4 + 6 = 15
Pink roses = 4

P(two pink roses) = (4/15)(3/14) = 2/35

Answer: 2/35

Solution 2

Guest Guest #3672
2 months ago

2/35, you can just multiply 4/15 x 3/14

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A flower vase has 5 white lilies, 4 pink roses, and 6 yellow carnations. One flower is chosen at random and given to a woman. Another flower is then chosen at random and given to a different woman. What is the probability that both flowers are pink roses?
Solution 1
 4/15·3/14 = 0.057 so it's b.) 0.057
Solution 2
5+4+6=15 2/15=.133333 or 13% or 2/15
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What is 4log1/2^w(2log1/2^u-3log1/2^v written as a single logarithm?
Solution 1
Given:

4log1/2^w (2log1/2^u-3log1/2^v)

Req'd:

Single logarithm = ?

Sol'n:

First remove the parenthesis,

4 log 1/2 (w) + 2 log 1/2 (u) - 3 log 1/2 (v)

Simplify each term,

Simplify the 4 log 1/2 (w) by moving the constant 4 inside the logarithm;
Simplify the 2 log 1/2 (u) by moving the constant 2 inside the logarithm;
Simplify the -3 log 1/2 (v) by moving the constant -3 inside the logarithm:

 log 1/2 (w^4)  + 2 log 1/2 (u) - 3 log 1/2 (v) 
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We have to use the product property of logarithms which is log of b (x) + log of b (y) = log of b (xy):

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Solution 2

Answer:

4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)=\log_{\frac{1}{2}}(\frac{w^4u^2}{v^3})

Step-by-step explanation:

Given : Expression  4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)

To write : As a single logarithm?

Solution :

4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)  

Remove parenthesis,

=4\log_{\frac{1}{2}}w+2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v  

Simplify each term by applying logarithmic property, a\log x=\log x^a

=\log_{\frac{1}{2}}w^4+\log_{\frac{1}{2}}u^2-\log_{\frac{1}{2}}v^3  

Use the product property of logarithms, \log_bx+\log_b y=\log_b (xy)

=\log_{\frac{1}{2}}w^4u^2-\log_{\frac{1}{2}}v^3  

Use the quotient property of logarithms, \log_bx-\log_b y=\log_b (\frac{x}{y})

=\log_{\frac{1}{2}}(\frac{w^4u^2}{v^3})  

Therefore,

4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)=\log_{\frac{1}{2}}(\frac{w^4u^2}{v^3})

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To solve this problem you must keep on min the information given in the problem shown above:

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Solution 2

Answer:

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Step-by-step explanation:

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