Mathematics: What is the probability getting 2 or 4 or 6 when a dice is tossed once?

Step-by-step explanation:For a function f(x) to satisfy the Mean Value Theorem, it must satisfy the following conditions:Be defined and continuous on the interval [a,b]Be differentiable on (a,b)There is at least one number c in the interval (a,b) (that is a c b) such that Given 1. f(x) is defined and continuous on [0,9] since a polynomial function is continuous everywhere.2. f(x) is differentiable in the interval (0,9) since a polynomial function is differentiable everywhere.3.By the mean value...

What is the probability getting 2 or 4 or 6 when

Ответ:

sarahsompayrac

13.10.2022

Mathematics

$c= \dfrac{9}{4}$

Step-by-step explanation:

For a function f(x) to satisfy the Mean Value Theorem, it must satisfy the following conditions:

Be defined and continuous on the interval [a,b]Be differentiable on (a,b)There is at least one number c in the interval (a,b) (that is a < c < b) such that $f'(c)=\dfrac{f(b)-f(a)}{b-a}$

Given $f(x)=\sqrt{x}$ in [0,9]$

1. f(x) is defined and continuous on [0,9] since a polynomial function is continuous everywhere.

2. f(x) is differentiable in the interval (0,9) since a polynomial function is differentiable everywhere.

$f(9)=\sqrt{9}=3\\f(0)=\sqrt{0}=0\\f'(x)=\dfrac{1}{2\sqrt{x} }$

3.By the mean value theorem:

$f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}\\\\\frac{1}{2\sqrt{c}} = \frac{f(9) - f(0)}{9-0} = \frac{3}{9}\\\\\frac{1}{2\sqrt{c}} = \frac{1}{3}\\\\$Cross multiply$\\\\2\sqrt{c}=3\\\\$Square both sides$\\\\(2\sqrt{c})^2=3^2\\\\4c=9\\\\c= \dfrac{9}{4}$

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What is the probability getting 2 or 4 or 6 when a dice is tossed once?

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