Answer:
See proof below
Step-by-step explanation:
Two triangles are said to be congruent if one of the 4 following rules is valid
- The three sides are equal
- The three angles are equal
- Two angles are the same and a corresponding side is the same
- Two sides are equal and the angle between the two sides is equal
Let's consider the two triangles ΔABC and ΔAED.
ΔABC sides are AB, BC and AC
ΔAED sides are AD, AE and ED
We have AE = AC and EB = CD
So AE + EB = AC + CD
But AE + EB = AB and AC+CD = AD
We have
AB of ΔABC = AD of ΔAED
AC of ΔABC = AE of ΔAED
Thus two sides the these two triangles. In order to prove that the triangles are congruent by rule 4, we have to prove that the angle between the sides is also equal. We see that the common angle is ∡BAC = ∡EAC
So triangles ΔABC and ΔAED are congruent
That means all 3 sides of these triangles are equal as well as all the angles
Since BC is the third side of ΔABC and ED the third side of ΔAED, it follows that
BC = ED Proved
Answer:
Vz = arccos(0.1673413807) = 80.37º
Step-by-step explanation:
The inverse function of the cosine is the arc cosine.
That means, that if we have that cos a = 1, a is the angle which has cosine equals 1, so a = arccos(1) = 0º.
In this problem, we have that:
cos(Vz) = 0.1673413807
Vz = arccos(0.1673413807) = 80.37º
Consider for a moment building a tree with all possible choices. The first row would contain 8 books (each possible choice for the first book). Then, the second row would contain 7 books for each book in the first row (the choices possible for the second book). Otherwise, it would be 8*7. Following this principal, we continue to multiply down to 1. 8*7*6*5*4*3*2*1. This is also known as 8!, a factorial.
Finally, 8! = 40,320
Answer:
a
The distribution of X is normal
b
c
Step-by-step explanation:
From the question we are told that
The population mean is 
The standard deviation is 
Generally given from the question that the amount of time spent alone by the population size is normally distributed then then the distribution of X (i.e the amount of time spent by the sample size (the one Mercurian)) will be normally distributed
Generally the probability that the child spend less than one hour in a day is mathematically represented as
Here 
So
From the z-table the value of
So
Generally the percentage of children that spends over 3.5 hours unsupervised is mathematically represented as
From the z-table the value of
So
