Where is the absolute total frequency located in a two-way frequency table

Mathematics

1 answer:

puteri [66]3 years ago

8 0

D. bottom right
Frequency is added here and found out!?

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Who can helppp ?????!

Elina [12.6K]

Answer:

well i do not know and would not know what those ovals are or whatever they are but the dude before is right .

Step-by-step explanation:

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3 years ago

Need help with AP CAL

anzhelika [568]

Answer: Choice C

$\displaystyle \frac{1}{2}\left(1 - \frac{1}{e^2}\right)$

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Explanation:

The graph is shown below. The base of the 3D solid is the blue region. It spans from x = 0 to x = 1. It's also above the x axis, and below the curve $y = e^{-x}$

Think of the blue region as the floor of this weirdly shaped 3D room.

We're told that the cross sections are perpendicular to the x axis and each cross section is a square. The side length of each square is $e^{-x}$ where 0 < x < 1

Let's compute the area of each general cross section.

$\text{area} = (\text{side})^2\\\\\text{area} = (e^{-x})^2\\\\\text{area} = e^{-2x}\\\\$

We'll be integrating infinitely many of these infinitely thin square slabs to find the volume of the 3D shape. Think of it like stacking concrete blocks together, except the blocks are side by side (instead of on top of each other). Or you can think of it like a row of square books of varying sizes. The books are very very thin.

This is what we want to compute

$\displaystyle \int_{0}^{1}e^{-2x}dx\\\\$

Apply a u-substitution

u = -2x

du/dx = -2

du = -2dx

dx = du/(-2)

dx = -0.5du

Also, don't forget to change the limits of integration

If x = 0, then u = -2x = -2(0) = 0
If x = 1, then u = -2x = -2(1) = -2

This means,

$\displaystyle \int_{0}^{1}e^{-2x}dx = \int_{0}^{-2}e^{u}(-0.5du) = 0.5\int_{-2}^{0}e^{u}du\\\\\\$

I used the rule that $\displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$ which says swapping the limits of integration will have us swap the sign out front.

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Furthermore,

$\displaystyle 0.5\int_{-2}^{0}e^{u}du = \frac{1}{2}\left[e^u+C\right]_{-2}^{0}\\\\\\= \frac{1}{2}\left[(e^0+C)-(e^{-2}+C)\right]\\\\\\= \frac{1}{2}\left[1 - \frac{1}{e^2}\right]$