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
6. area of parallelogram is b x h. the base is 3, the height is 2. the height IS NOT 2.2. THAT IS THE SLANT HEIGHT NOT THE HEIGHT
3x²-11=x+13
3x²-x-24=0
(3x+8)(x-3)=0
x=3 or -8/3
☺☺☺☺
3. Three and four tenths. 3+0.4=3.4
4. Two and fifty-one hundredths. 2+0.51=2.51
5. 8/10
6. 0.05
7. 46/100
8. 0.6
9. 9/10
10. 0.35
1 and 2 are correct btw
Answer:
The answer is C.
Step-by-step explanation:
m=y²-y¹/x²-x¹
m=(-12)-13/2-7
m=-25/-5
m=5
