The scores for a 25 - point quiz are listed below arranged from least to greatest. 7, 7, 12, 13, 15, 16, 16, 16, 18, 19, 20, 20, 20, 21, 21, 22, 23, 23, 24 Using the following guidelines , create a histogram for the class.

a. Determine the range of the sample
b. Determine the number of intervals by: Finding the square root of n. Round to the next higher whole number.
c. Determine the interval width: Width = Range/# Intervals. Round to the next higher whole number.
d. Create the categories.
e. Plot the data

Step-by-step explanation:

Given

The above data

Solving (a): The range:

Where

So:

Solving (b): Number of intervals.

From the given data:

Solving (c): Interval width

This is calculated as:

--- round up

Solving (d): Create the categories

Based on the calculated parameters above, the categories are: 7 - 10, 11 - 14, 15 - 18, 19 - 22 and 23 - 26

Solving (e): The histogram

First construct the frequency table

Intervals ---- Frequency --- Midpoint

7 - 10  ---------    2  ----------------  8.5

11 - 14    ---------- 2 ----------------- 12.5

15 - 18    --------  5 ----------------- 16.5

19 - 22    -------- 7 ------ ----------- 21.5

23 - 26 ---------- 3 ----------------- 24.5

The midpoint is calculated by calculating the mean of the intervals.

For instance:

For class 7 - 10, the midpoint is: (7+10)/2 = 17/2= 8.5

This is applied to other classes too

The midpoint is then plotted against the frequency.

See attachment

a

step-by-step explanation:

a

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}