What Is the answer submitted in one !!!!

How do I find the Q1, Q3, and IQR of a median that is a decimal?

bija089 [108]

Step 1: Put the numbers in order.
1,2,5,6,7,9,12,15,18,19,27Step 2: Find the median (How to find a median).
1,2,5,6,7,9,12,15,18,19,27Step 3: Place parentheses around the numbers above and below the median.
Not necessary statistically–but it makes Q1 and Q3 easier to spot.
(1,2,5,6,7),9,(12,15,18,19,27)Step 4: Find Q1 and Q3
Q1 can be thought of as a median in the lower half of the data. Q3 can be thought of as a median for the upper half of data.
(1,2,5,6,7), 9, ( 12,15,18,19,27). Q1=5 and Q3=18.Step 5: Subtract Q1 from Q3 to find the interquartile range.
18-5=13.
Hope this helped:)
-BRIEMODEE:)

3 0

3 years ago

How did the graph of f(x) = x^2 and g(x) = 3/4 x^2 relate?

nirvana33 [79]

g(x) = $\frac{3}{4} f(x)$ or g(x) is 3/4 times of f(x) , F(x) and g(x) have common solution or intersecting point in the graph parabola at x=0 i.e. in origin and x = $\frac {4}{3}$ .

Step-by-step explanation:

We have a function f(x) = $x^{2}$ and another function , g(x) = $\frac{3}{4} x^{2}$ . In the graph of y = $x^{2}$ , the point (0, 0) is called the vertex. The vertex is the minimum point in a parabola that opens upward. In a parabola that opens downward, the vertex is the maximum point.

Graphing y = (x - h)2 + k , where h = 0 & k = 0

Function g(x) can be formed with compression in function f(x) by a factor of 3/4 , i.e. g(x) = $\frac{3}{4} f(x)$ or g(x) is 3/4 times of f(x).Domain and range of f(x) and g(x) are same ! Although structure of both functions is same the only difference is g(x) is compressed vertically by a factor 3/4. Both are graph of a parabola with vertex at (0,0). Also, F(x) and g(x) have common solution or intersecting point at x=0 i.e. in origin.

7 0

3 years ago

Use a tape diagram to represent and find the value of 1/2 /1/3

NARA [144]

Answer:

Step-by-step explanation:

Take the last number in the fraction ( the denominator ) and make a segment that long. Than, take the first number in the fraction ( the numerator ) and highlight, bolden, color, or do whatever needs to be done to make sure that the numerator's value stand out.

The link at the bottom is an example...

7 0

2 years ago

Water drips into a circular puddle such that the radius of the puddle, in centimeters, at time t, in seconds, is given by the eq

bulgar [2K]

Part A

Given that the puddle is circular in shape and that the radius of the puddle, in centimeters, at time t, in seconds, is given by the equation $r(t)=\sqrt{t}$ .

Then the area of the puddle is given by the area of a circle = $Area=\pi r^2$
But, given that $r(t)=\sqrt{t}$ , then $A(t)=\pi (r(t))^2=\pi(\sqrt{t})^2=\pi t$

Therefore, the equation for the area of the puddle as a function of t is given by $A(t)=\pi t$

Part B

The average rate of change of a function f(x) between x = a and x = b is given by $\frac{f(b)-f(a)}{b-a}$ .

Thus, the average rate of change of the area of the puddle with respect to time between t = 0 and t = 16 is given by $\frac{A(16)-A(0)}{16-0} = \frac{16\pi-0}{16} = \frac{16\pi}{16} =\pi$

Therefore, the average rate of change of the area of the puddle with respect to time between t = 0 and t = 16 is π.

Part C

The area of the puddle with respect to the radius is given by $A(r)=\pi r^2$

Given that $r(t)=\sqrt{t}$ , thus when t = 0, $r(0)=\sqrt{0}=0$ and when t = 16, $r(16)=\sqrt{16}=4$

Thus, the average rate of change of the area of the puddle with respect to the radius between r = 0 and r = 4 is given by

$\frac{A(4)-A(0)}{4-0} = \frac{\pi(4)^2-\pi(0)^2}{4} = \frac{16\pi}{4} =4\pi$

Therefore, the average rate of change of the area of the puddle with respect to the radius between t = 0 and t = 16 is 4π.

Part D

The circumference of a circle is given by $C=2\pi r$

Thus, the radius of the puddle in terms of circumference is given by $r= \frac{C}{2\pi}$

Thus, the area of the puddle with respect to the circumference, C, of the puddle is given by $A(C)=\pi\left( \frac{C}{2\pi} \right)^2= \frac{1}{4\pi} C^2$

Since, $C=2\pi r$ and $r(t)= \sqrt{t}$ , thus when t = 0, r = 0 and C = 0; when t = 16, r = 4 and C = 8π.

Thus the area of the puddle with respect to the circumference, C, of the puddle between C = 0 and C = 8π is given by $\frac{A(8\pi)-A(0)}{8\pi-0} = \frac{ \frac{(8\pi)^2}{4\pi}- \frac{(0)^2}{4\pi} }{8\pi} = \frac{ \frac{64\pi^2}{4\pi} }{8\pi} = \frac{16\pi}{8\pi} =2$

Therefore, the average rate of change of the area of the puddle with respect to the circumference of the puddle between t = 0 and t = 16 is 2.

5 0

3 years ago

I went to the store to buy candy for my family. If I bought a total of 42 pieces of candy and I give my family of five the most

Katyanochek1 [597]

You will have 2 pieces left for yourself.
because 42 divided by 5 will not give you a whole number you set aside two because 40 is divisible by 5 leaving you with the whole number of 8 pieces for each family member and two remaining for yourself.

4 0

3 years ago