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Does there exist an algorithm I can use to take two random multidigit integers and averaging them digit by digit? If you have the numbers $a$, $b$ with the base $10$ representations \begin{align} a &= (d_{m-1} d_{m-2} \dotsm d_0)_{10} = \sum_{k=0}^{m-1} d_k \, 10^k \\ b &= (D_{M-1} D_{M-2} \dotsm D_0)_{10} = \sum_{k=0}^{M-1} D_k \, 10^k ...

3

If we assume that $$v'(t)=a(t),\quad t\in [a,b],$$ then there is no difference, since $$\int_{a}^{b} a(t) dt=\int_{a}^{b} v'(t) dt=v(b)-v(a),$$ giving $$a_{avg}=\frac{1}{b-a} \int_{a}^{b} a(t) dt=\frac{v(b)-v(a)}{b-a}=m_{secant}.$$

2

If $x$ is not in the list, we will use $25$ comparisons. For $x$ in the list, we assume the elements in the list are distinct. We also assume, not necessarily reasonably, that $x$ is equally likely to be the first element, the second, and so on. So the average number of comparisons is $\frac{1}{25}(1+2+\cdots+25)$, which simplifies to $13$. The average ...

2

Not too hard actually! Once I realized everything was by definition positive... $$0 \leq (a - b)^2$$ $$\dfrac{-(a - b)^2}{2} \leq 0$$ $$\dfrac{-a^2 + 2ab - b^2}{4} \leq 0$$ $$\dfrac{a^2 + 2ab + b^2}{4} - \dfrac{a^2 + b^2}{2} \leq 0$$ $$0 \leq (\dfrac{a + b}{2})^2 \leq \dfrac{a^2 + b^2}{2}$$ Since both sides are positive, we can take a square root ...

2

In certain situations, especially many situations involving rates and ratios, the harmonic mean provides the truest average. https://en.wikipedia.org/wiki/Harmonic_mean

1

You made an arithmetic error when you calculated $\frac{1\cdot 20+0\cdot 5}{25}=\frac{20+0}{25}=\frac{4}{5}=0.8\neq 0.83$ Indeed, making the correction, you should arrive at: $\frac{25-25\cdot 0.8}{25\cdot 0.8} = \frac{25-20}{20}=\frac{5}{20}=0.25$ as expected.

1

You are right. The question is incomplete. Some more information is needed to solve the question.

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According to the law of cosines given $a,b,\gamma$ to be respectively two sides ($a \ge b$) and the angle in between you get that $$c^2 = a^2+b^2-2ab\cos \gamma$$ so the average of the three sides is $$\frac{a+b+c}{3} = \frac{a+b+\sqrt{a^2+b^2-2ab\cos \gamma}}{3}$$ Now note that the minimum of the average is when $\cos \gamma = 1 \Rightarrow$ ...

1

The average speed is total distance divided by total time. The percentages are distance. Let the total distance be $D$, though it will divide out. She went $0.1D$ at $56$ mph. How long did that take (it will include a factor $D$)? Do the same for the other two legs, add up the times, divide into $D$.

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The total average speed is the total distance divided by the total time. We don't know the total length of the trip, but let's assume for simplicity that the whole distance is $100$ miles (we'll see that value doesn't matter in the end). Let's call the single distances $d_1,d_2,d_3$ where $d_1+ d_2 + d_3 = 100$. I'll denote the speeds by $v_i$ and the times ...

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Your second approach is wrong. Consider for example this situation: The company has a $99.9\%$ probability of having $0$ daily profit and a probability of $0.01\%$ of having a profit of one million. Now the average profit will be $0.999 \cdot 0 + 0.001 \cdot 10^6 = 10000$. But if you simply take the average of the maximal and the minimal possible profit, ...

1

The average of $10$, $99$, and $101$ is $\dfrac{10+99+101} 3 = 70$, and that is not the same as the average of the smallest and largest of these numbers, which is $\dfrac{10+101} 2 = 55.5$. That is because $99$ is a lot bigger than the average of the smallest and largest ones. It is not generally true that the average of the numbers in a list is the same ...

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You have height $h$ as a function of time $h=f(t)$ The rate of change of height is the velocity so take the derivative of height with respect to time to get a function for velocity $v$. To get the average of $v$ from time $t=2$ to a final time $t=T$ use the integral $\frac{1}{T-2}\int_{2}^{T}v dt$

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