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dusya [7]
3 years ago
12

17. Which of the following measures can be calculated for qualitative data? a. Mean b. Median c. Mode d. All of these

Mathematics
1 answer:
VLD [36.1K]3 years ago
6 0

Answer: C. Mode

Step-by-step explanation:

Qualitative data applies to observable data such as reoccurring patterns (mode)

Quantitative data applies to numbers and numerical calculations (Mean, Median)

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Robert is purchasing some packs of chicken from the supermarket. The first pack weighs 45 ounces. The second pack weighs 3 pound
Dmitry [639]

Answer: 3 pounds

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
PLZ SHOW ALL WORK
lara31 [8.8K]

Part I
We have the size of the sheet of cardboard and we'll use the variable "x" to represent the length of the cuts. For any given cut, the available distance is reduced by twice the length of the cut. So we can create the following equations for length, width, and height.
width:  w = 12 - 2x
length: l = 18 - 2x
height: h = x

Part II
v = l * w * h
v = (18 - 2x)(12 - 2x)x
v = (216 - 36x - 24x + 4x^2)x
v = (216 - 60x + 4x^2)x
v = 216x - 60x^2 + 4x^3
v = 4x^3 - 60x^2 + 216x

Part III
The length of the cut has to be greater than 0 and less than half the length of the smallest dimension of the cardboard (after all, there has to be something left over after cutting out the corners). So 0 < x < 6

Let's try to figure out an x that gives a volume of 224 in^3. Since this is high school math, it's unlikely that you've been taught how to handle cubic equations, so let's instead look at integer values of x. If we use a value of 1, we get a volume of:
v = 4x^3 - 60x^2 + 216x
v = 4*1^3 - 60*1^2 + 216*1
v = 4*1 - 60*1 + 216
v = 4 - 60 + 216
v = 160

Too small, so let's try 2.
v = 4x^3 - 60x^2 + 216x
v = 4*2^3 - 60*2^2 + 216*2
v = 4*8 - 60*4 + 216*2
v = 32 - 240 + 432
v = 224

And that's the desired volume.
So let's choose a value of x=2.
Reason?
It meets the inequality of 0 < x < 6 and it also gives the desired volume of 224 cubic inches.
3 0
3 years ago
Consider this expression:
Tema [17]
8(4a+2b)
The equivalent is 32a+16b (B)

4 0
2 years ago
Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term x^2, which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form f(x)=ax^2+bx+c is given by the expression: x_v=\frac{-b}{2a}

Since in our case a=1 and b=5, we get that the x-position of the vertex is: x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the x_v=-\frac{5}{2}. Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

5 0
3 years ago
Answer the questions below. When you are finished, submit this assignment to your teacher by the due date for full credit. Total
Andrews [41]
Jill had a total score of 40 on 13 quizzes, so her average was
  40/13 ≈ 3.08

_____
1 quiz @ 1 = total of 1
3 quizzes @ 2 = total of 6
4 quizzes @ 3 = total of 12
4 quizzes @ 4 = total of 16
1 quiz @ 5 = total of 5
total score = 1 + 6 + 12 + 16 + 5 = 40
total quizzes = 1 + 3 + 4 + 4 + 1 = 13
8 0
3 years ago
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