Answer:
AO bisects BAC - Needed
AO is perpendicular to BC - Needed
Point O is the midpoint of BC - Not needed
AO is an altitude - Not needed
Step-by-step explanation:
In ΔAOB,
- <BOA = 90° (since AO is perpendicular to BC
)
- <BAO = <BAC (since AO bisects BAC)
In ΔAOC,
- <COA = 90° (since AO is perpendicular to BC
)
- <CAO = <BAC (since AO bisects BAC)
In ΔAOB and ΔAOC,
- <BOA = <COA (since both are 90°)
- <BAO = CAO (since AO bisects BAC)
- AO=AO
So, we can conclude ΔAOB ≅ ΔAOC (<em>ASA property</em>) and hence the base angles <ABO and < ACO of the isosceles triangle ABC are also congruent.
Solution :
a). The probability that the student will the 1st question after the 4th attempt.
P (correct in the 4th attempt)
=
= 0.01171875
b). The probability that the student will 3 questions after 10 total attempts.
P( X = 3) for X = B in (n = 10, p = 0.75)
=
= 0.0031
c). The mean and the standard deviation for the number of attempts up to when the students gets all the questions correct is :
There are = 6 success, p = 0.75.
Therefore, this is a case of a negative binomial distribution.
= 8
So,
= 1.6330
Answer:
s=2r
Step-by-step explanation:
Solution:
Given;
2q-r=p
r=q-s
or,q=r+s
p=q+2r
To prove: s=2r
Now,
2q-r=p
or,2(r+s)-r=q+2r
or,2r+2s-r=r+s+2r
or,r+2s=3r+s
or,s=2r
PROVED
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Answer:
a) 0.25x
b) (0.25x+5)*3
Step-by-step explanation: