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olchik [2.2K]
3 years ago
13

You roll a pair of six-sided number cubes, numbered 1 through 6. What is the probability of rolling two odd numbers? (Enter your

probability as a fraction
Mathematics
1 answer:
lbvjy [14]3 years ago
8 0

Answer:

1/4 or 25%

Step-by-step explanation:

The probability for rolling an odd number is 50% or  \frac{1}{2} because there are 3 odd numbers, (1, 3, 5) and 6 total outcomes (1, 2, 3, 4, 5, 6). \frac{3}{6} =\frac{1}{2}

The probability for rolling two odd numbers in a row is

\frac{1}{2} *\frac{1}{2} =\frac{1}{4}

So, the probability of rolling an odd number twice is 25% or  \frac{1}{4}.

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The graph shows the amount of a medicine min milligrams, remaining in a patient's body hours after receiving an injection . The
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<h3>(a) The factor at which the medicine decreased, in 1.5 hour</h3>

From the graph, we have the following ordered pairs

(x,y) = (0,270) and (1.5,80)

Given that:

y = ab^x

At point (0,270), we have:

ab^0=270

a= 270

At point (1.5,80), we have:

ab^{1.5} = 80

Substitute 270 for a

270b^{1.5} = 80

Divide both sides by 270

b^{1.5} = 0.2963

Take 1.5th root of both sides

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<h3>(b) The factor at which the medicine decreased in the first half hour, and the first hour</h3>

In (a), we have:

The decay factor (b) to be 0.44

This represents the factor at which the medicine decreased throughout.

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In (a), we have:

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So, the exponential equation is:

y =270 * 0.44^x

In terms of m and h, we have:

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Read more about exponential equations at:

brainly.com/question/11832081

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2 years ago
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