Split the triangle into a 30-60-90 triangle. Since you have LM from that triangle, you can figure out the other sides. Then, using the leg from that triangle, the one next to it is a 45-45-90 triangle, meaning it's isosceles.
KM = 30 + 10√3
KL = 10√6
(sorry i'm not too sure if my "reasons" for statement/reason is correct)
Consider the attached diagram. Segment KO is an altitude of the triangle, so KO ⊥ LM. ∠L has measure 180° -105° -30° = 45°, so ΔKOL is an isosceles right triangle.
If we let segment LO have measure 1, then KO also has measure 1 and KL has measure √(1²+1²) = √2 by the Pythagorean theorem.
ΔKMO is half of an equilateral triangle, so KM has measure 2, and MO has measure √(2²-1²) = √3 by the Pythagorean theorem.
Then the ratio of KM to LM is 2:(1+√3) and the ratio of KL to LM is √2:(1+√3). That is, ...
... KL = LM×(√2)/(1+√3) = (20√3)(√2)/(1 +√3)
... KL = (20√6)/(√3 +1) = (20√6)(√3 -1)/(3 -1) . . . . . with denominator rationalized
... KL = 30√2 -10√6 ≈ 17.9315
and
... KM = LM×2/(1+√3) = KL×√2
... KM = (30√2 -10√6)√2
... KM = 60 -20√3 ≈ 25.3590
== == == == == ==
The Law of Sines tells you ...
... KL/sin(M) = KM/sin(L) = LM/sin(K)
Then ...
... KL = sin(M)·LM/sin(K) = sin(30°)·20√3/sin(105°) ≈ 17.9315
... KM = sin(L)·LM/sin(K) = sin(45°)·20√3/sin(105°) ≈ 25.3590
Answer:
Option c
Step-by-step explanation:
A system of equations are given to us. And we need to solve them . The given system is
We numbered the equations here . Now put the value of equation 1 in equation 2 that is substituting y = 2x - 3 in eq. 2 .
We got two values of x as 0 & 2 . Alternatively substituting these values we have ,
Thefore the required answer is ,
Answer:
a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.
b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.
c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.
Step-by-step explanation:
We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.
This means that , in which 60 is the number of miles.
(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?
, and thus,
This probability is:
In which
0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.
(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?
Binomial distribution.
6 blizzard days means that
Each of these days, 0.6592 probability of no accident on this stretch, which means that .
This probability is P(X = 2). So
0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.
(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?
Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:
80 miles:
This means that . So
The probability of no damaging accidents is P(X = 0). So
0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.