Using the information given, it is found that the class width for this frequency distribution table is of 1.
In this problem, these following classes are given:
0 – 1 14
2 – 3 1
4 – 5 8
6 – 7 12
8 – 9 12
The classes not given, which are 1 - 2, 3 - 4 and 5 - 6, have values of 0.
The <u>difference between the bounds of the classes is of 1</u>, thus, the class width is of 1.
A similar problem is given at brainly.com/question/24701109
Answer:
58°
Step-by-step explanation:
5.6*1 2/5 = 5 3/5 * 1 2/5 =28/5 * 7/5 = 196/25 = 784/100 = 7.84 or 7 21/25
It is equal to 0.8 five star please?
