A die with 8 sides is rolled.what is the probability of rolling a number less than 7

Mathematics

1 answer:

larisa86 [58]3 years ago

3 0

We want that number to be 1 2 3 ....6

the probability is = 6/8 we simplify by 2=3/4

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S_A_V [24]

Taking $y=y(x)$ and differentiating both sides with respect to $x$ yields

$\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0$

Solving for the first derivative, we have

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y$

Differentiating again gives

$\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0$

Solving for the second derivative, we have

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}$

Now, when $x=1$ and $y=2$ , we have

$\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63$

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**HELP!**

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