Answer:
don't know
Step-by-step explanation:
Answer:
a) 9000 ft-lb
b) 9000 ft-lb
Step-by-step explanation:
Equation for the work:
The equation for the work done is given by:
In which F is the force needed and d is the distance. The time does not interfere on the work.
A 300-lb gorilla climbs a tree to a height of 30 ft
This means that 
Thus
The work done in both cases is of 9000 ft-lb, as the work is not a function of time.
The steps on the construction of a segment bisector by paper folding, and label the midpoint M is given below.
<h3>What are the steps of this construction?</h3>
1. First, one need to open a Compass so that it is said to be more than half the length of the said segment.
2. Without altering it, with the aid of the compass, do draw an art above and also below the said line segment from one of the segment endpoints.
3. Also without altering it and with use the compass, do draw another pair of arts from the other and points. One arc will be seen above the segment and the other or the second arc will be seen below.
4. Then do draw the point of intersection that is said to exist between the pair of arts below the line segment and also in-between the pair of arts as seen below the line segment
5. Lastly, do make use of a straight edge to link the intersection points between the both pair of arts.
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The first decimal point is tenth. 4.88
when rounding, we just look at the number right after the number you want to round to. it's 8 right? since 8 is closer to 10 than 0,
we round it to 4.9
(4.8 + 0.10)
Answer:
The relative frequency is found by dividing the class frequencies by the total number of observations
Step-by-step explanation:
Relative frequency measures how often a value appears relative to the sum of the total values.
An example of how relative frequency is calculated
Here are the scores and frequency of students in a maths test
Scores (classes) Frequency Relative frequency
0 - 20 10 10 / 50 = 0.2
21 - 40 15 15 / 50 = 0.3
41 - 60 10 10 / 50 = 0.2
61 - 80 5 5 / 50 = 0.1
81 - 100 <u> 10</u> 10 / 50 = <u>0.2</u>
50 1
From the above example, it can be seen that :
- two or more classes can have the same relative frequency
- The relative frequency is found by dividing the class frequencies by the total number of observations.
- The sum of the relative frequencies must be equal to one
- The sum of the frequencies and not the relative frequencies is equal to the number of observations.
