A plane leaves airport A and travels 570 miles to airport B on a bearing of Upper N 32 degrees Upper E. The plane later leaves a
irport B and travels to airport C 390 miles away on a bearing of Upper S 73 degrees Upper E. Find the distance from airport A to airport C to the nearest tenth of a mile.
1 answer:
Answer:
769.46 miles
Step-by-step explanation:
Please have a look at the photo attached in this answer
The missing part of the angle is 32 degrees, so the total angle of B is:
32 + 73 = 105 degrees.
Let x is the distance from airport A to airport C to the nearest tenth of a mile, we use cosine law to find x:
=
- 2(570)(390)cos(105)
<=> = 592070
<=> x = = 769.46 miles
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Answer:
0.03125 = 3.125% probability of obtaining 5 heads from 5 coin flips.
Step-by-step explanation:
For each coin flip, there are only two possible outcomes. Either it is heads, or it is tails. The probabilities for each flip are independent from each other. So we use the binomial probability distribution to solve this question.
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.
In this problem we have that:
For each coin toss, heads and tails are equally as likely, so
What is the probability of obtaining 5 heads from 5 coin flips?
This is P(X = 5) when n = 5.
So
0.03125 = 3.125% probability of obtaining 5 heads from 5 coin flips.