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What is Cumulative Frequency Curve or the Ogive in Statistics

December 3, 2020December 3, 2020 by Raju

What is Cumulative Frequency Curve or the Ogive in Statistics

First we prepare the cumulative frequency table, then the cumulative frequencies are plotted against the upper or lower limits of the corresponding class intervals. By joining the points the curve so obtained is called a cumulative frequency curve or ogive.
There are two types of ogives :

Less than ogive : Plot the points with the upper limits of the class as abscissae and the corresponding less than cumulative frequencies as ordinates. The points are joined by free hand smooth curve to give less than cumulative frequency curve or the less than Ogive. It is a rising curve.
Greater than ogive : Plot the points with the lower limits of the classes as abscissa and the corresponding Greater than cumulative frequencies as ordinates. Join the points by a free hand smooth curve to get the “More than Ogive”. It is a falling curve.

When the points obtained are joined by straight lines, the picture obtained is called cumulative frequency polygon.
Less than ogive method:
To construct a cumulative frequency polygon and an ogive by less than method, we use the following algorithm.
Algorithm
Step 1 :     Start with the upper limits of class intervals and add class frequencies to obtain the cumulative frequency distribution.
Step 2 :     Mark upper class limits along X-axis on a suitable scale.
Step 3 :     Mark cumulative frequencies along Y-axis on a suitable scale.
Step 4 :     Plot the points (xi, fi) where xi is the upper limit of a class and fi is corresponding cumulative frequency.
Step 5 :     Join the points obtained in step 4 by a free hand smooth curve to get the ogive and to get the cumulative frequency polygon join the points obtained in step 4 by line segments.

More than ogive method:
To construct a cumulative frequency polygon and an ogive by more than method, we use the following algorithm.
Algorithm
Step 1 :     Start with the lower limits of the class intervals and from the total frequencysubtract the frequency of each class to obtain the cumulative frequency distribution.
Step 2 :     Mark the lower class limits along X-axis on a sutiable scale.
Step 3 :   Mark the cumulative frequencies along Y-axis on a suitable scale.
Step 4 :     Plot the points (xi, fi) where xi is the lower limit of a class and fi is corresponding cumulative frequency.
Step 5 :     Join the points obtained in step 4 by a free hand smooth curve to get the ogive and to get the cumulative frequency polygon join these points by line segments

Read More:

Cumulative Frequency Curve or the Ogive Example Problems with Solutions

Example 1: Draw a less than ogive for the following frequency distribution :

I.Q.	Frequency
60 – 70	2
70 – 80	5
80 –90	12
90 – 100	31
100 – 110	39
110 – 120	10
120 – 130	4

Find the median from the curve.
Solution: Let us prepare following table showing the cumulative frequencies more than the upper limit.

Class interval (I. Q)	Frequency (f)	Cumulative frequency
60 – 70	2	2
70 – 80	5	2 + 5 = 7
80 –90	12	2 + 5 + 12 = 19
90 – 100	31	2 + 5 + 12 + 31 = 50
100 – 110	39	2 + 5 + 12 + 31 + 39 = 89
110 – 120	10	2 + 5 + 12 + 31 + 39 + 10 = 99
120 – 130	4	2 + 5 + 12 + 31 + 39 + 10 + 4 = 103

Less than ogive :
I.Q. is taken on the x-axis. Number of students are marked on y-axis.
Points (70, 2), (80, 7), (90, 19), (100, 50), (110, 89), (120, 99), (130, 103), are plotted on graph paper and these points are joined by free hand. The curve obtained is less than ogive.
What is Cumulative Frequency Curve or the Ogive in Statistics 1
The value \(\frac{N}{2}\) = 51.5 is marked on y-axis and from this point a line parallel to x-axis is drawn. This line meets the curve at a point P. From P draw a perpendicular PN to meet x-axis at N. N represents the median.
Here median is 100.5.
Hence, the median of given frequency distribution is 100.5

Example 2: The following table shows the daily sales of 230 footpath sellers of Chandni Chowk.

Sales in Rs.	No. of sellers
0 – 500	12
500 – 1000	18
1000 – 1500	35
1500 – 2000	42
2000 – 2500	50
2500 – 3000	45
3000 – 3500	20
3500 – 4000	8

Locate the median of the above data using only the less than type ogive.
Solution: To draw ogive, we need to have a cumulative frequency distribution.

Sales in Rs.	No. of sellers	Less than type cumulative frequency
0 – 500	12	12
500 – 1000	18	30
1000 – 1500	35	65
1500 – 2000	42	107
2000 – 2500	50	157
2500 – 3000	45	202
3000 – 3500	20	222
3500 – 4000	8	230

Less than ogive :
Seles in Rs. are taken on the y-axis and number of sellers are taken on x-axis. For drawing less than ogive, points (500, 12), (1000, 30), (1500, 65), (2000, 107), (2500, 157), (3000, 202), (3500, 222), (4000, 230) are plotted on graph paper and these are joined free hand to obtain the less than ogive.
What is Cumulative Frequency Curve or the Ogive in Statistics 2
The value \(\frac{N}{2}\) = 115 is marked on y-axis and a line parallel to x-axis is drawn. This line meets the curve at a point P. From P draw a perpendicular PN to meet x-axis at median. Median = 2000.
Hence, the median of given frequency distribution is 2000.

Example 3: Draw the two ogives for the following frequency distribution of the weekly wages of (less than and more than) number of workers.

Weekly wages	Number of workers
0 – 20	41
20 – 40	51
40 – 60	64
60 – 80	38
80 – 100	7

Hence find the value of median.
Solution:

Weekly wages	Number of workers	C.F (less than)	C.F (More than)
0 – 20	41	41	201
20 – 40	51	92	160
40 – 60	64	156	109
60 – 80	38	194	45
80 – 100	7	201	7

Less than curve :
Upper limits of class intervals are marked on the x-axis and less than type cumulative frequencies are taken on y-axis. For drawing less than type curve, points (20, 41), (40, 92), (60, 156), (80, 194), (100, 201) are plotted on the graph paper and these are joined by free hand to obtain the less than ogive.
What is Cumulative Frequency Curve or the Ogive in Statistics 3
Greater than ogive
Lower limits of class interval are marked on x-axis and greater than type cumulative frequencies are taken on y-axis. For drawing greater than type curve, points (0, 201), (20, 160), (40, 109), (60, 45) and (80, 7) are plotted on the graph paper and these are joined by free hand to obtain the greater than type ogive. From the point of intersection of these curves a perpendicular line on x-axis is drawn. The point at which this line meets x-axis determines the median. Here the median is 42.652.

Example 4: Following table gives the cumulative frequency of the age of a group of 199 teachers.
Draw the less than ogive and greater than ogive and find the median.

Age in years	Cum. Frequency
20 – 25	21
25 – 30	40
30 – 35	90
35 – 40	130
40 – 45	146
45 – 50	166
50 – 55	176
55 – 60	186
60 – 65	195
65 – 70	199

Solution:

Age in years	Less than cumulative frequency	Frequency	Greater than type
20 – 25	21	21	199
25 – 30	40	19	178
30 – 35	90	50	159
35 – 40	130	40	109
40 – 45	146	16	69
45 – 50	166	20	53
50 – 55	176	10	33
55 – 60	186	10	23
60 – 65	195	9	13
65 – 70	199	4	4

Find out the frequencies by subtracting previous frequency from the next frequency to get simple frequency. Now we can prepare the greater than type frequency. Ages are taken on x-axis and number of teachers on y-axis.
Less than ogive :
Plot the points (25, 21), (30, 40), (35, 90), (40, 130), (45, 146), (50, 166), (55, 176), (60, 186), (65, 195), (70, 199) on graph paper. Join these points free hand to get less than ogive.
Greater than ogive :
Plot the points (20, 199), (25, 178), (30, 159), (35, 109), (40, 69), (45, 53), (50, 33), (55, 23), (60, 13), (65, 4) on graph paper. Join these points freehand to get greater than ogive. Median is the point of intersection of these two curves.
What is Cumulative Frequency Curve or the Ogive in Statistics 4
Here median is 37.375.

Example 5: Following is the age distribution of a group of students. Draw the cumulative frequency polygon, cumulative frequency curve (less than type) and hence obtain the median value.

Age	Frequency
5 – 6	40
6 – 7	56
7 – 8	60
8 – 9	66
9 – 10	84
10 – 11	96
11 – 12	92
12 – 13	80
13 – 14	64
14 – 15	44
15 – 16	20
16 – 17	8

Solution: We first prepare the cumulative frequency table by less then method as given below :

Age	Frequency	Age less than	Cumulative frequency
5 – 6	40	6	40
6 – 7	56	7	96
7 – 8	60	8	156
8 – 9	66	9	222
9 – 10	84	10	306
10 – 11	96	11	402
11 – 12	92	12	494
12 – 13	80	13	574
13 – 14	64	14	638
14 – 15	44	15	682
15 – 16	20	16	702
16 – 17	8	17	710

Other than the given class intervals, we assume a class 4-5 before the first class interval 5-6 with zero frequency.
Now, we mark the upper class limits (including the imagined class) along X-axis on a suitable scale and the cumulative frequencies along Y-axis on a suitable scale.
Thus, we plot the points (5, 0), (6, 40), (7, 96), (8, 156), (9, 222), (10, 306), (11, 402), (12, 494), (13, 574), (14, 638), (15, 682), (16, 702) and (17, 710).
These points are marked and joined by line segments to obtain the cumulative frequency polygon shown in Fig.
What is Cumulative Frequency Curve or the Ogive in Statistics 5
In order to obtain the cumulative frequency curve, we draw a smooth curve passing through the points discussed above. The graph (fig) shows the total number of students as 710. The median is the age corresponding to \(\frac{N}{2}\,\, = \,\,\frac{{710}}{2}\) = 355 students. In order to find the median, we first located the point corresponding to 355th student on Y-axis. Let the point be P. From this point draw a line parallel to the X-axis cutting the curve at Q. From this point Q draw a line parallel to Y-axis and meeting X-axis at the point M. The x-coordinate of M is 10.5 (See Fig.). Hence, median is 10.5.
What is Cumulative Frequency Curve or the Ogive in Statistics 6

Example 6: The following observations relate to the height of a group of persons. Draw the two type of cumulative frequency polygons and cumulative frequency curves and determine the median.

Height in cms	140–143	143–146	146–149	149–152	152–155	155–158	158–161
Frequency	3	9	26	31	45	64	78
Height in cms	161–164	164–167	167–170	170–173	173–176	176–179	179–182
Frequency	85	96	72	60	43	20	6

Solution: Less than method : We first prepare the cumulative frequency table by less than method as given below :

Height in cms	Frequency	Height less than	Frequency
140–143	3	143	3
143–146	9	146	12
146–149	26	149	38
149–152	31	152	69
152–155	45	155	114
155–158	64	158	178
158–161	78	161	256
161–164	85	164	341
164–167	96	167	437
167–170	72	170	509
170–173	60	173	569
173–176	43	176	612
176–179	20	179	632
179–182	6	182	638

Other than the given class intervals, we assume a class interval 137-140 prior to the first class interval 140-143 with zero frequency.
Now, we mark the upper class limits on X-axis and cumulative frequency along Y-axis on a suitable scale.
We plot the points (140, 0), (143, 3), (146, 12), (149, 38), (152, 69), (155, 114), (158, 178), (161, 256),
(164, 341), (167, 437), (170, 509), (173, 569), (176, 612),.(179, 632) and 182, 638).
What is Cumulative Frequency Curve or the Ogive in Statistics 7
These points are joined by line segments to obtain the cumulative frequency polygon as shown in fig. and by a free hand smooth curve to obtain an ogive by less than method as shown in fig.

More than method : We prepare the cumulative frequency table by more than method as given below :
Other than the given class intervals, we assume the class interval 182-185 with zero frequency.
Now, we mark the lower class limits on X-axis and the cumulative frequencies along Y-axis on suitable scales to plot the points (140, 638), (143, 635), (146, 626), (149, 600), (152, 569), (155, 524), (158, 460), (161, 382), (164, 297), (167, 201), (170, 129), (173, 69), (176, 26) and (179, 6). By joining these points by line segments, we obtain the more than type frequency polygon as shown in fig. By joining these points by a free hand curve, we obtain more than type cumulative frequency curve as points by a free hand curve, we obtain more than type cumulative frequency curves as shown in fig.
We find that the two types of cumulative frequency curves intersect at point P. From point P perpendicular PM is drawn on X-axis. The value of height corresponding to M is 163.2 cm. Hence, median is 163.2 cm.

What is a Grouped Frequency Distribution Table

December 3, 2020December 2, 2020 by Raju

What is a Grouped Frequency Distribution Table

There are 3 methods for calculation of mean :

Direct Method
Assumed mean deviation method
Step deviation method.

1. Direct Method for Calculation of Mean
What is a Grouped Frequency Distribution Table 1
According to direct method

2. Assumed Mean Method
Arithmetic mean = \(a + \frac{{\sum\limits_{i = 1}^n {{f_i}{d_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\)
Note : The assumed mean is chosen, in such a manner, that

It should be one of the central values.
The deviation are small.
One deviation is zero.

Working Rule :
Step 1 :       Choose a number ‘a’ from the central values of x of the first column, that will be our assumed mean.
Step 2 :      Obtain deviations d_i by subtracting ‘a’ from x_i. Write down hese deviations against the corresponding frequencies in the third column.
Step 3 :      Multiply the frequencies of second column with corresponding deviations d_i in the third column to prepare a fourth column of f_id_i.
Step 4 :      Find the sum of all the entries of fourth column to obtain ∑f_id_i and also, find the sum of all the frequencies in the second column to obtain ∑f_i.

Read More:

3. Step Deviation Method
Deviation method can be further simplified on dividing the deviation by width of the class interval h. In such a case the arithmetic mean is reduced to a great extent.
Mean (\(\bar x\)) = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}} \times h\)
Working Rule :
Step-1 :     Choose a number ‘a’ from the central values of x(mid-values)
Step-2 :    Obtain u_i = \(\frac{{{x_i} – a}}{h}\)
Step-3 :    Multiply the frequency f_i with the corresponding u_i to get f_iu_i.
Step-4 :    Find the sum of all f_iu_ii.e., ∑f_iu_i
Step-5 :     Use the formula = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}} \times h\) to get the required mean.

Grouped Frequency Distribution Table Example Problems with Solutions

Example 1:

Mid-values	2	3	4	5	6
Frequencies	49	43	57	38	13

Find the mean by direct method.

Solution:

Mid Values	frequencies (f_i)	f_ix_i
2	49	98
3	43	129
4	57	228
5	38	190
6	13	78
Total	N = Σf_i = 50	Σf_ix_i = 2750

Mean = \(\frac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}\) = \(\frac{{723}}{{200}}\) = 3.615

Example 2: Find the mean of the following frequency distribution :

Class Interval	Frequency
10-30	90
30-50	20
50-70	30
70-90	20
90-110	40

Solution:

Class Interval	f	Mid value (x)	f × x
10-30	90	20	1800
30-50	20	40	800
50-70	30	60	1800
70-90	20	80	1600
90-110	40	100	4000
	Σf = 200		Σfx = 10000

Mean = \(\frac{{\sum {f}{x}}}{{\sum {f}}}\) = \(\frac{{10000}}{{200}}\) = 50

Example 3: A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants	0 – 2	2 – 4	4 – 6	6 – 8	8 – 10	10 – 12	12 – 14
No. of houses	1	2	1	5	6	2	3

Which method did you use for finding the mean and why ?
Solution:

Number of plants	Number of houses (f)	Mid value (x)	f × x
0-2	1	1	1
2-4	2	3	6
4-6	1	5	5
6-8	5	7	35
8-10	6	9	54
10-12	2	11	22
12-14	3	13	39
	Σf = 20		Σfx = 162

Mean = \(\frac{{\sum {f}{x}}}{{\sum {f}}}\) = \(\frac{{162}}{{20}}\) = 8.1

Example 4: Calculate the mean for the following distribution:

Variable	5	6	7	8	9
Frequency	4	8	14	11	3

Solution:
What is a Grouped Frequency Distribution Table 3
∴ Mean = \(\frac{{\sum f\,x}}{{\sum f}} = \frac{{281}}{{40}}\) = 7.025

Example 5: Find the mean of the following frequency distribution :
What is a Grouped Frequency Distribution Table 4
Solution:

Mean = \(\frac{{\sum f\,x}}{{\sum f}} = \frac{{4930}}{{150}} = 32.8\overline 6\) or 32.87 (approx.)

Example 6: Find the mean of the following distribution by direct method.

Class interval	0 – 10	11 – 20	21 – 30	31 – 40	41 – 50
Frequency	3	4	2	5	6

Solution:
What is a Grouped Frequency Distribution Table 6
Mean = \(\frac{{\sum f\,x}}{{\sum f}} = \frac{{578.5}}{{20}}\) = 28.9

Example 7: For the following distribution, calculate mean using all the suitable methods.

Size of Item	1 – 4	4 – 9	9 – 16	16 – 27
Frequency	6	12	26	20

Solution:
What is a Grouped Frequency Distribution Table 7
Mean = \(\frac{{\sum f\,x}}{{\sum f}} = \frac{{848}}{{64}}\) = 13.25

Example 8: The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.

Expenditure (in rupees)	100-150	150-200	200-250	250-300	300-350	350-400	400-450	450-500
Frequency	24	40	33	28	30	22	16	7

Solution: Let assumed mean = 275
What is a Grouped Frequency Distribution Table 8
\(\bar x = a + \frac{{\Sigma {f_i}{d_i}}}{{\Sigma {f_i}}}\) = 275 + \(\frac{{ – 1750}}{{200}}\) = Rs 266.25

Example 9: Calculate the arithmetic mean of the following distribution :

Class Interval	Frequency
0 – 50	17
50 –100	35
100 –150	43
150–200	40
200– 250	21
250– 300	24

Solution: Let assumed mean = 175 i.e. a = 175
What is a Grouped Frequency Distribution Table 9
Now , a = 175
\(\bar x = a + \frac{{\Sigma {f_i}{d_i}}}{{\Sigma {f_i}}}\) = 175 + \(\frac{{ – 4750}}{{180}}\)
= 175 – 26.39 = 148.61 approx.

Example 10: Calculate the arithmetic mean of the following frequency distribution :

Class interval	50– 60	60–70	70–80	80–90	90– 100
Frequency	8	6	12	11	13

Solution: Let assumed mean = 75 i.e., a = 75
What is a Grouped Frequency Distribution Table 10
a = 75, Σf_id_i= 150, Σf_i = 50
Mean \(\bar x = a + \frac{{\Sigma {f_i}{d_i}}}{{\Sigma {f_i}}}\) = 75 + \(\frac{{ 150}}{{50}}\) = 78

Example 11: Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute	Frequency
65– 68	2
68–71	4
71–74	3
74–77	8
77– 80	7
80– 83	4
83– 86	2

Solution: Let assumed mean a = 75.5
What is a Grouped Frequency Distribution Table 11
Mean = \(a + \frac{{\Sigma fd}}{{\Sigma f}} = 75.5 + \frac{{12}}{{30}}\) = 75.5 + 0.4 = 75.9

Example 12: To find out the concentration of SO₂ in the air (in parts per million, i.e.ppm), the data was collected for 30 localities in a certain city and is presented below :
What is a Grouped Frequency Distribution Table 12
Find the mean concentration of SO₂ in the air.
Solution: Let the assumed mean a = 0.10.

By step deviation method
Mean = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}}\) × h
= 0.10 + \(\frac{{–1}}{{30}} \times 0.04\)
= 0.10 – 0.0013
= 0.0987
= 0.099 ppm

Example 13: The weekly observation on cost of living index in a certain city for the year 2004–2005 are given below. Compute the mean weekly cost of living index.
What is a Grouped Frequency Distribution Table 14
Solution: Let assumed mean be 1750 i.e., a = 1750

By step deviation method
Mean (\(\bar x\)) = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}}\) × h
= 1750 + \(\frac{{ – 45}}{{52}} \times 100\)
= 1750 – 86.54
= 1663.46
Hence, the mean weekly cost of living index
= 1663.46

Example 14: Find the mean marks from the following data by step deviation method
What is a Grouped Frequency Distribution Table 16
Solution: Let assumed mean = 55 ⇒ a = 55

Here, a = 55, h = 10,
Σf_i = 85, Σf_iu_i = –56
Mean (\(\bar x\)) = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}}\) × h
h = 55 + \(\frac{{ – 56}}{{85}} \times 10\)
= 55 – 6.59 = 48.41
Hence, mean mark = 48.41.

Example 15: Find the mean age of 100 residents of a colony from the follwing data :
What is a Grouped Frequency Distribution Table 18
Solution: Let assumed mean a = 35

Here, a = 35, h = 10
\(\bar x\) = a + \(\frac{{\Sigma {f_i}{u_i}}}{{\Sigma {f_i}}}\) × h
⇒ \(\bar x\) = 35 + \(\frac{{ – 40}}{{100}} \times 10\) = 31
Hence, the mean age = 31 years

Example 16: The following distribution show the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18.00. Find the missing frequency f.
What is a Grouped Frequency Distribution Table 20
Solution: we have,

Mean \(\bar x\) = \(\frac{{\Sigma fx}}{{\Sigma f}}\) ⇒   18 = \(\frac{{752 + 20f}}{{44 + f}}\)
⇒   18 (44 + f) = 752 + 20f
⇒   752 + 20f = 792 + 18f
⇒ 2f = 40
⇒      f = 20
Hence, the missing frequency is 20.

Example 17: The arithmetic mean of the following frequency distribution is 50. Find the value of p.
What is a Grouped Frequency Distribution Table 22
Solution:

Mean \(\bar x\) = \(\frac{{\Sigma fx}}{{\Sigma f}}\)  ⇒ 50 = \(\frac{{5160 + 30P}}{{92 + P}}\)
⇒   50 (92 + P) = 5160 + 30 P
⇒   4600 + 50 P = 5160 + 30P
⇒   20 P = 560
⇒   P = 28

Example 18: The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f1 and f2 :
What is a Grouped Frequency Distribution Table 24
Solution:

30 + f₁ + f₂ = 50 ⇒ f₁ + f₂ = 20    ….(1)
Mean = \(\frac{{\Sigma fx}}{{\Sigma f}}\) ⇒ 62.8 = \(\frac{{2060 + 30{f_1} + 70{f_2}}}{{50}}\)
⇒ 62.8 = \(\frac{{206 + 3{f_1} + 7{f_2}}}{5}\)
⇒ 206 + 3f₁ + 7f₂ = 314
⇒   3f₁ + 7f₂ = 108                     ….(2)
3f₁ + 3f₂ = 60                            ….(3)
[Multiplying (1) by 3]
On Subtracting (3) from (2), we get
4f₂ = 48 ⇒ f₂ = 12
Putting f₂ = 12 in (1), we get
f₁ = 8

Glencoe Algebra 1 Solutions Chapter 13 Statistics Exercise 13.5

December 22, 2017March 3, 2017 by Dattu

Glencoe Algebra 1 Solutions Chapter 13 Statistics Exercise 13.5

Glencoe Algebra 1 Chapters Answer Key

Glencoe Algebra 1 Solutions Chapter 13 Statistics Exercise 13.5

Answer 1AA.
Glencoe-Algebra-1-answers-Statistics-13.5-1aa

Answer 1CU.

Answer 2CU.

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Answer 3AA.

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Glencoe-Algebra-1-answers-Statistics-13.5-56mys
Answer 57MYS.

Glencoe Algebra 1 Solutions Chapter 13 Statistics Exercise 13.4

December 22, 2017March 3, 2017 by Dattu

Glencoe Algebra 1 Solutions Chapter 13 Statistics Exercise 13.4

Glencoe Algebra 1 Chapters Answer Key

Glencoe Algebra 1 Solutions Chapter 13 Statistics Exercise 13.4

Answer 1CU.
Glencoe-Algebra-1-answers-Statistics-13.4-1cu

Answer 1PQ.

Answer 2CU.

Answer 2PQ.

Answer 3CU.

Answer 3PQ.

Answer 4CU.

Answer 4PQ.

Answer 5CU.

Answer 5PQ.

Answer 6CU.

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Answer 10CU.

Answer 11PA.

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Glencoe Algebra 1 Solutions Chapter 13 Statistics Exercise 13.3

December 22, 2017March 3, 2017 by Dattu

Glencoe Algebra 1 Solutions Chapter 13 Statistics Exercise 13.3

Glencoe Algebra 1 Chapters Answer Key

Glencoe Algebra 1 Solutions Chapter 13 Statistics Exercise 13.3

Answer 1CU.
Glencoe-Algebra-1-answers-Statistics-13.3-1cu

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Glencoe-Algebra-1-answers-Statistics-13.3-12pa

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Glencoe-Algebra-1-answers-Statistics-13.3-27pa

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Glencoe-Algebra-1-answers-Statistics-13.3-39mys
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