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kaheart [24]
3 years ago
15

The estimated value of the integral from 0 to 2 of x cubed dx , using the trapezoidal rule with 4 trapezoids is

Mathematics
1 answer:
bulgar [2K]3 years ago
4 0
The integral is approximated by the sum,

\displaystyle\int_0^2f(x)\,\mathrm dx\approx\sum_{n=0}^4\frac12\times\frac{f(x_n)+f(x_{n+1})}2=\frac14\sum_{n=0}^3(f(x_n)+f(x_{n+1}))

where f(x)=x^3 and x_n=\dfrac12n, giving you

\displaystyle\frac14\sum_{n=0}^3\bigg(\left(\frac n2\right)^3+\left(\frac{n+1}2\right)^3\bigg)
\displaystyle\frac1{32}\sum_{n=0}^3(n^3+(n+1)^3)
\displaystyle\frac1{32}\sum_{n=0}^3(2n^3+3n^2+3n+1)

Faulhaber's formulas make short work of computing the sum. You have

\displaystyle\sum_{n=0}^k1=k+1
\displaystyle\sum_{n=0}^kn=\frac{k(k+1)}2
\displaystyle\sum_{n=0}^kn^2=\frac{k(k+1)(2k+1)}6
\displaystyle\sum_{n=0}^kn^3=\frac{k^2(k+1)^2}4

which gives

\displaystyle\frac1{16}\sum_{n=0}^3n^3+\frac3{32}\sum_{n=0}^3n^2+\frac3{32}\sum_{n=0}^3n+\frac1{32}\sum_{n=0}^31
\displaystyle\frac{36}{16}+\frac{42}{32}+\frac{18}{32}+\frac4{32}
\implies\displaystyle\int_0^2x^3\,\mathrm dx\approx\frac{17}4=4.25
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single die is rolled twice. Find the probability of rolling an oddodd number the first time and a number greater than 33 the sec
givi [52]

Answer:

\frac{1}{4}

Step-by-step explanation:

A single die is rolled twice, we have to find the probability of rolling an odd number in first throw and a number greater than 3 in the second throw.

a) Rolling an Odd number in first throw

A die has total 6 possible outcomes, out of which 3 are odd numbers i.e. 1,3 and 5

So, total number of possible outcomes = 6

Total Favorable outcomes (Odd numbers) = 3

Probability is defined as the ratio of favorable outcomes to total number of outcomes. So,

The probability of rolling an odd number would be = \frac{3}{6} = \frac{1}{2}

b) Rolling a number greater than 3 in second throw

Here again total possible outcomes = 6

Favorable outcomes (Numbers greater than 3 are 4, 5 and 6) = 3

So,

The probability of rolling a number greater than 3 =  \frac{3}{6} = \frac{1}{2}

These two events(rolls) are independent of each other, so the overall probability of both events occurring would be the product of individual probabilities.

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Probability of rolling an odd number the first time and a number greater than 3 the second time = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

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2 years ago
The amount of time, in minutes, that a woman must wait for a cab is uniformly distributed between zero and 12 minutes, inclusive
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Answer:

P(X\leq x) =\frac{x-a}{b-a}, a \leq x \leq b

And using this formula we have this:

P(X

Then we can conclude that the probability that that a person waits fewer than 11 minutes is approximately 0.917

Step-by-step explanation:

Let X the random variable of interest that a woman must wait for a cab"the amount of time in minutes " and we know that the distribution for this random variable is given by:

X \sim Unif (a=0, b =12)

And we want to find the following probability:

P(X

And for this case we can use the cumulative distribution function given by:

P(X\leq x) =\frac{x-a}{b-a}, a \leq x \leq b

And using this formula we have this:

P(X

Then we can conclude that the probability that that a person waits fewer than 11 minutes is approximately 0.917

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3 years ago
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Answer:

c. $24,435

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