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if a distribution of raw scores were plotted and then the scores were transformed to z scores, would the shape of the distributi

nadya68 [22]

<span>•<span>If the raw score is transformed into a z-score, however, the value of the z-score tells exactly where the score is located relative to all the other scores in the distribution. </span></span>

3 0

3 years ago

Think about the function f(x) = 3 - 2x

Angelina_Jolie [31]

f(0) means the function output when the input is x = 0. This is the same as saying the y value when x = 0.

f(x) = 3-2x

f(0) = 3-2(0)

f(0) = 3

The point (0,3) is on the graph. This is the y intercept which is where the graph crosses the y axis. The y intercept always occurs when x = 0.

So in other words, the special name for f(0) is the y intercept.

6 0

4 years ago

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The length of a rectangle is 4 cm less than it’s width. If the area of the rectangle is 165 cm^2. What are the dimensions of the

Monica [59]

Answer:

The dimensions are 11 cm by 15 cm

Step-by-step explanation:

Area of a rectangle = (length)(width).

Let L represent the length and W the width.

Since L = W - 4, we have

Area of rectangle = (W - 4)W = 165 cm^2.

Thus, w^2 - 4W - 165 = 0. We can solve for W using the quadratic formula:

a = 1, b = -4 and c = -165. Thus,

-(-4) ± √ [ (-4)² - 4(1)(-165) ]

W = ---------------------------------------

2

4 ± √ [16 + 660 ]

W = ------------------------------

2

4 ± √ [ 676 ] 4 ± 26

W = ----------------------- = W = ------------- = W = 2 ± 13

2 2

Thus, W is 2 + 13 = 15. It cannot be negative, so we discard 2 - 13 = -11.

If the width, W, is 15, then the length is 4 less, or L = 11.

The dimensions are 11 cm by 15 cm

6 0

4 years ago

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

4 years ago

Which of the following shows the polynomial below written in descending order?

forsale [732]

To write the polynomial in descending order, arrange the terms from the highest to the lowest degree. Simply base on the exponents of the variable x. You have 3, 12, 1 and 7. In descending order, that would be: 12, 7, 3 and then 1. The answer is D.

4 0

3 years ago

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