Advanced Placement (AP), 31.08.2020 01:01 becky6962

PLEASE HELP! DUE IN 10 MIN Describe how the process of sampling, RGB pixels, and binary sequences work together to display a digital color image. 20 POINT REWARDD

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Answer from: isbella29

When an image is sampled, it means that the image is transformed from analog to digital.

For this to be done, the image must undergo two processes.

Binary sequenceRGB pixels

Binary sequence

This is the first sampling process, where the analog images get sampled through series of 0s and 1s.

Once the binary sequencing is done, the next step is the RGB pixels.

The RGB pixels:

This stage of sampling receives from the binary sequence, the analog images.

The RGB pixels use a digital color pixel (as numbers) to represent the RGB code (i.e. Red, Green and Blue).

When the images are sampled, the digital colors are then displayed on the screen.

Read more about the sampling at:

link

Answer from: flameftw2005

Explanation:

"Pixel" is short for "picture element."

An image can be described in digital form by any of several methods. When RGB pixels are used, each pixel in the image is coded as three 8-bit binary values. Typically, a binary value of 00000000 corresponds to none of the associated color (red, green, or blue), and a binary value of 11111111 corresponds to the maximum amount of the associated color. The amount of the color used to render the pixel is proportional to the binary value.*

A pixel, coded as three 8-bit color values, is a sample of the image. The proportion of the image represented depends on the resolution, typically expressed as a number of horizontal or vertical pixels per image. An image may be 4096 pixels wide and 3072 pixels high, for example, so would consist of about 12.6 million samples. Some coding schemes associate a linear measurement with the size of a pixel: 160 pixels per inch means the the image just described would be rendered 25.6 inches wide by 19.2 inches high. Depending on the rendering device, a pixel may or may not be square. (Resolution in one direction may differ from that in another direction.)

In a camera, the incoming light is sampled by circuits sensitive to the different colors. The output from those circuits is digitized and combined appropriately so that the RGB values for a given pixel represent the same location in the image.

In a rendering device, the various color-producing processes are aligned (registered) so that the red, green, and blue values are combined appropriately at the proper spot in the image. There may be additional processing to take into account the color-rendering characteristics of the device used. That is, there may be some non-linearity between the RGB values for a pixel and the amount of light or pigment produced by the rendering device. The purpose of this is to ensure that the colors produced are as close as possible to the colors in the original scene captured by the camera.

In short, a camera or computer represents the pixels in a scene by binary values representing the red, green, and blue (RGB) components of the color at a point. Those binary values are interpreted by a rendering device (display or printer) to reproduce the original color as nearly as possible.

Answer from: Quest

The answer is (d) 7/2======we have $f(x) = \begin{cases} x & \text{ for } x \le 0 \\ x + 1 & \text{ for } x \ \textgreater \ 0 \end{cases}$ we also have an integral $\displaystyle\int_{-2}^1 x\, f(x)\; dx$ notice how function f splits up at x = 0. there's a left piece and a right pieceso we'll need to do the same thing with our integral: split it at x = 0we use a property of integrals that allows us to split up and sum: $\displaystyle\int_{-2}^1 x\, f(x)\; dx = \displaystyle\int_{-2}^0 x\, f(x)\; dx + \displaystyle\int_{0}^1 x\, f(x)\; dx$ notice that for the piecewise function f, the function has the equationf(x) = x when x ≤ 0. so for the integral $\int_{-2}^0 x\, f(x)\; dx$ , f(x) = x so it changes into $\begin{aligned} \int_{-2}^0 x\, f(x)\; dx & = \int_{-2}^0 x \cdot x\; dx \\ & = \int_{-2}^0 x^2\; dx \\ & = \left[\frac{x^3}{3}\right]_{-2}^0 & & \text{\footnotesize antidifferentiate} \\ & = 0 - \frac{(-2)^3}{3} \\ & = \frac{8}{3} \end{aligned}$ for the piecewise function f, the function has the equationf(x) = x+1 when x > 0. for the integral $\int_{0}^1 x\, f(x)\; dx$ , we have $\begin{aligned} \int_{0}^1 x\, f(x)\; dx & = \int_{0}^1 x(x+1)\; dx \\ & = \int_{0}^1 (x^2+x)\; dx \\ & = \left[ \frac{x^3}{3} + \frac{x^2}{2}\right]_0^1 \\ & = \frac{1}{3} + \frac{1}{2} - 0 = \frac{5}{6} \end{aligned}$ therefore, back to our splitted integral: $\begin{aligned} \displaystyle\int_{-2}^1 x\, f(x)\; dx & = \displaystyle\int_{-2}^0 x\, f(x)\; dx + \displaystyle\int_{0}^1 x\, f(x)\; dx \\ & = \frac{8}{3} + \frac{5}{6} \\ & = \frac{7}{2} \end{aligned}$
Can anyone me on this ap calc question? ?

Can anyone me on this ap calc question? ?

Answer from: Quest

Yep this is right you

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