WILL GIVE BRAINLIEST! In any boxplot, what percent of the data is above the 3rd quartile?

1 answer:

5 0

75% because a third quartile is 3/4 which is 75% which if you break each fourth down it would be 25% so then you multiply 25%x3 to get 75%

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Vertically stretched by a factor of 2, then translated<br> 4 units left and 1 unit down

drek231 [11]

A vertical stretch of scale factor 2, followed by a translation of 4 units left and 1 unit down is written as:

g(x) = 2*f(x + 4) - 1

<h3>How to write the given transformation?</h3>

For a general function f(x), a vertical stretch of scale factor K is written as:

g(x) = K*f(x).

<u><em>Horizontal translation:</em></u>

For a general function f(x), a horizontal translation of N units is written as:

g(x) = f(x + N).

If N is positive, the shift is to the left.
If N is negative, the shift is to the right.

<u><em>Vertical translation:</em></u>

For a general function f(x), a vertical translation of N units is written as:

g(x) = f(x) + N.

If N is positive, the shift is upwards.
If N is negative, the shift is downwards.

So, if we start with a function f(x) and we stretch it vertically with a scale factor of 2, we get:

g(x) = 2*f(x)

Then we translate it 4 units left:

g(x) = 2*f(x + 4)

Then we translate 1 unit down:

g(x) = 2*f(x + 4) - 1

This is the equation for the transformation.

If you want to learn more about transformations, you can read:

brainly.com/question/4289712

6 0

2 years ago

PLEASE HELP!!!!!!!dgbdgdbhdndcn

bogdanovich [222]

Problem 1)

AC is only perpendicular to EF if angle ADE is 90 degrees

(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88

Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle

Triangle AED is acute (all 3 angles are less than 90 degrees)

So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF

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Problem 2)

a) The center is (2,-3)

The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2

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b) The radius is 3 and the diameter is 6

From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2

where
h = 2
k = -3
r = 3

so, radius = r = 3
diameter = d = 2*r = 2*3 = 6

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c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.

Some points on the circle are

A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)

Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.