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Mice21 [21]
3 years ago
14

2. What is the probability of spinning an A on the first spinner and an odd number

Mathematics
1 answer:
pychu [463]3 years ago
5 0

Answer: <em>1/10</em>

Step-by-step explanation: These two events are independent, meaning the outcome of the first event doesn't affect the outcome of the second.

To find the probability of independent events, first, find the

probability of each event, then multiplies the probabilities.

The probability of spinning an A on the first spinner is 1/5

since there is 1 favorable outcome and 5 total outcomes.

The probability of spinning an odd number on the second spinner

is 3/6 which simplifies to 1/2 since there are 3 favorable outcomes,

spinning a 1, 3, or another 3 since these are odd numbers.

So now we have 1/5 × 1/2 and when multiplying fractions,

we multiply across the numerators and denominators to get 1/10.

So the probability of spinning an A, then an odd number is 1/10.

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Answer:

Step-by-step explanation:

Given the definite integral \int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx, we to evaluate it. Using integration by substitution method.

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= \int\limits {\dfrac{x^4}{u^5} } \, \dfrac{du}{-10x^4}

= \dfrac{-1}{10} \int\limits {\dfrac{du}{u^5} }  \\\\= \dfrac{-1}{10} \int\limits {{u^{-5}du }  \\= \dfrac{-1}{10} [{\frac{u^{-5+1}}{-5+1}]  \\\\= \dfrac{-1}{10} ({\frac{u^{-4}}{-4})\\\\

= \dfrac{u^{-4}}{40} \\\\\\= \dfrac{1}{40u^4} +C

substitute u = 1-2x⁵ into the result

= \dfrac{1}{40(1-2x^5)^4} +C

Hence\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx = \dfrac{1}{40(1-2x^5)^4} +C

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