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Salsk061 [2.6K]
2 years ago
6

HELP ASAP!!!

Mathematics
2 answers:
Veseljchak [2.6K]2 years ago
7 0
I don't know. This seems tricky. But I do know that the probability would be 1 divided by the probability. So A and D is Incorrect. Theres never really a 50-50 chance with Locker doors so I'm gonna go with C.

I'm not guaranteed this is the answer. But I do Hope it helps. =)

icang [17]2 years ago
3 0
The problem is complicated since there are a lot of variables to consider, however I do know that it's probably D.

We know that the first digit is more than 5. And the second we are guessing the probability for which is if it will be more than 5.

Now I can't say I do know the answer or how to solve it, but I know it's definitely not 1/2 or 1/3. I wouldn't say it's B. either because you cannot include 5, which if it was true for that equation then we're saying including the first digit we have 5 chances, however it only goes up to 9 because we don't count zero.

Hope this helped in some way.

- Kathryn Irene
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I don't speak Romanian, but the closest translation for this suggests you're trying to compute

\displaystyle \int x^3 \ln(x)^2 \, dx

Integrate by parts:

\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du

where

u = ln(x)²   ⇒   du = 2 ln(x)/x dx

dv = x³ dx   ⇒   v = 1/4 x⁴

\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx

Integrate by parts again:

\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'

where

u' = ln(x)   ⇒   du' = dx/x

dv' = x³ dx   ⇒   v' = 1/4 x⁴

\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx

So, we have

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C

\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C

\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}

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2 years ago
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