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If AM=BM, then point M is a middle point of side AB.
1. Consider triangle AML. You know that AM=ML, then this triangle is isosceles and AL is its base. The angles adjacent to the base of isosceles triangle are conruent, this means that
2. Consider lines ML and AC. The angle bisector AL is transversal. Since alternate interior angles you have that lines ML and AC are parallel. This means that ML is a middle line of triangle and 2ML=AC. Also you know that AC=2AL. This gives you that ML=AL. Now ML=AL and ML=AM gives you that triangle AML is equilateral.
3. In equilateral triangle AML all angles are congruent and have measures 60°. Thus, m∠AML=m∠MLA=m∠LAM=60°.
4. AL is angle bisector, then m∠MAL=m∠LAC=60° and m∠BAC=m∠MAL+m∠LAC=120°.
5. Consider ΔBML, it is isosceles, because BM=ML and m∠BML=180°-m∠AML=180°-60°=120°. Then,
6. Consider triangle ABC. In this triangle m∠A=120°, m∠B=30°, then
m∠C=180°-m∠A-m∠B=180°-120°-30°=30°.
Answer: m∠A=120°, m∠B=m∠C=30°.
Answer:
The number of hits would follow a binomial distribution with and
.
The probability of finding hits is approximately
(or equivalently, approximately
.)
The mean of the number of hits is approximately . The variance of the number of hits is approximately
(not the same number as the mean.)
Step-by-step explanation:
There are possible passwords in this set. (Approximately two billion possible passwords.)
Each one of the randomly-selected passwords would have an approximately
chance of matching one of the users' password.
Denote that probability as :
.
For any one of the randomly-selected passwords, let
denote a hit and
denote no hits. Using that notation, whether a selected password hits would follow a bernoulli distribution with
as the likelihood of success.
Sum these 's and
's over the set of the
randomly-selected passwords, and the result would represent the total number of hits.
Assume that these randomly-selected passwords are sampled independently with repetition. Whether each selected password hits would be independent from one another.
Hence, the total number of hits would follow a binomial distribution with trials (a billion trials) and
as the chance of success on any given trial.
The probability of getting no hit would be:
.
(Since is between
and
, the value of
would approach
as the value of
approaches infinity.)
The mean of this binomial distribution would be:.
The variance of this binomial distribution would be:
.