I believe the answer is:
1 13889/25000
Give me a min if this one is wrong. "Thank" if i got it right
Answer:
y=68
Step-by-step explanation:
Answer:
Class Boundary = 1 between the sixth and seventh classes.
Step-by-step explanation:
Lengths (mm) Frequency
1. 140 - 143 1
2. 144 - 147 16
3. 148 - 151 71
4. 152 - 155 108
5. 156 - 159 83
6. 160 - 163 18
7. 164 - 167 3
The class boundary between two classes is the numerical value between the starting value of the higher class, which is 164 for the 7th class in this case, and the ending value of the class of the lower class, which is 163 for the 6th class in this case.
Therefore the class boundary between the sixth and seventh classes
= 164 - 163 = 1
Therefore Class Boundary = 1.
It can be seen that class boundary for the frequency distribution is 1.
If we take the difference between the lower limits of one class and the lower limit of the next class then we will get the class width value.
Therefore, Class width,
w = lower limit of second class - lower limit of first class
= 144 - 140
= 4
B is not a geometric sequence
Answer:
378 ft²
Step-by-step explanation:
Given:
The Width of the wall = 28'
The Length of the wall = 10'
The Height of the gable = 7'
Now, the width of the gable will be equal to the width of the wall = 28'
therefore,
Area of the wall = Area of the rectangle section of wall + area of the gable
Total area of the wall = ( 28' × 10' ) + (
)
or
Total area of the wall = 280 + 98 = 378 ft²