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3 years ago

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A standard number cube with the numbers 1 through 6 is rolled. Find the probability of rolling a number less than 5. Find the pr

obability of not rolling a number less than 5.

1 answer:

kvv77 [185]3 years ago

5 0

Probability of rolling a 5 is 1 out of 6
probability of rolling a number under 5 is 4 out of 6

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valentina_108 [34]

The fundamental theorem of calculus tells us that

$\vec v(t)=\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du$

$\vec r(t)=\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du$

So we have

$\vec v(t)=(\vec\imath-\vec\jmath-2\,\vec k)+\displaystyle\int_0^t(u^2\,\vec\imath-2u^{1/2}\,\vec\jmath+e^{3u}\,\vec k)\,\mathrm du$

$\vec v(t)=(\vec\imath-\vec\jmath-2\,\vec k)+\left(\dfrac{u^3}3\,\vec\imath-\dfrac{4u^{3/2}}3\,\vec\jmath+\dfrac{e^{3u}}3\,\vec k\right)\bigg|_0^t$

$\vec v(t)=\dfrac{t^3+3}3\,\vec\imath-\dfrac{4t^{3/2}+3}3\,\vec\jmath+\dfrac{e^{3t}-7}3\,\vec k$

and

$\vec r(t)=(2\,\vec\imath+\vec\jmath-\vec k)+\displaystyle\int_0^t\left(\frac{u^3+3}3\,\vec\imath-\frac{4u^{3/2}+3}3\,\vec\jmath+\frac{e^{3u}-7}3\right)\,\mathrm du$

$\vec r(t)=(2\,\vec\imath+\vec\jmath-\vec k)+\left(\dfrac{u^3+12u}{12}\,\vec\imath-\dfrac{8u^{5/2}+15u}{15}\,\vec\jmath+\dfrac{e^{3u}-21u}9\,\vec k\right)\bigg|_0^t$

$\vec r(t)=\dfrac{t^3+12t+24}{12}\,\vec\imath-\dfrac{8t^{5/2}+15t-15}{15}\,\vec\jmath+\dfrac{e^{3t}-21t-10}9\,\vec k$

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Step-by-step explanation:

18+25=43

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3 years ago

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slavikrds [6]

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Step-by-step explanation:

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