For a), this is clearly a given as it is literally to the right of where it says “Given:”
For b), since ON bisects ∠JOH, this means that it splits it into two separate angles - JON and HON, which are similar due to that bisects mean that it splits it equally into two halves
For c), since NO is the same thing as NO, it is equal to itself
For d), since AAS (angle-angle-side) congruence states that if there are two angles that are congruent (proved in a) and b) ) as well as that a side is congruent (proved in c) ), two triangles are congruent
For e), since two triangles are congruent, every side must have one side that it matches up to in the other triangle. As the opposite side of angle H is JO and the opposite side of angle J is OH, and ∠J=∠H, those two are congruent. As JN and HN are the two sides left, they must be congruent.
Feel free to ask further questions!
Answer:
i. 18 miles per hour is an outlier
ii. the outlier decreases the mean speed
Step-by-step explanation:
An outlier in a given data is one of the values that is far greater or lesser compared to others. It affect the mean and standard deviation of a given data significantly.
From the given data, 18 is far too small compared to other values. This is certainly an outlier. This would affect the mean speed by decreasing the value.
An interquartile range is a measure of differences among data by dividing a set of given data into quartile. Increasing the value of the outlier would increase the interquartile range.
k/jkjln/kjln/j
Answer:
yes
Step-by-step explanation:
pic above
The +2 goes on the Y-Axis… From
the 2 count up 3 times and go to the right 2 times
Answer:
V = 20.2969 mm^3 @ t = 10
r = 1.692 mm @ t = 10
Step-by-step explanation:
The solution to the first order ordinary differential equation:
Using Euler's method
Where initial droplet volume is:
Hence, the iterative solution will be as next:
- i = 1, ti = 0, Vi = 65.45
- i = 2, ti = 0.5, Vi = 63.88
- i = 3, ti = 1, Vi = 62.33
We compute the next iterations in MATLAB (see attachment)
Volume @ t = 10 is = 20.2969
The droplet radius at t=10 mins
The average change of droplet radius with time is:
Δr/Δt = 
The value of the evaporation rate is close the value of k = 0.08 mm/min
Hence, the results are accurate and consistent!
