Answer:
A= 154, B=22
Step-by-step explanation:
if meterorite A is 7 times bigger than B, then our equation would be 7B+B=176. This gives us B=22. Then you multiply 22 times 7 to get your A value.
I´d say "d" is the distance from the eye to the wall.
Now substracting 1.2-1 you´ll get the distance of the wall of the smallest triangle = 0.2 And you do 1.5-0.2= 0.3 that´s the distance of the wall of the other triangle. Then you solve everything with Pitagoras theorem. You have 2 rectangle triangles.
B+alfa=45°
tan^-1(0.2/d)=B
tan^-1(1.3/d)=alfa
THEN:
tan^-1(0.2/d)+tan^-1(1.3/d)=45°
Now you have 3 ecs and 3 variables.
alfa,B and "d"
Answer:
x=-6 is an extraneous solution
Step-by-step explanation:
we have
squared both sides
solve the quadratic equation by graphing
The solution is x=-6 and x=-1
see the attached figure
<u><em>Verify each solution</em></u>
substitute the value of x in the original expression
For x=-6
----> is not true
so
Is an extraneous solution
For x=-1
----> is true
so
Is the solution
Answer:
- The probability that overbooking occurs means that all 8 non-regular customers arrived for the flight. Each of them has a 56% probability of arriving and they arrive independently so we get that
P(8 arrive) = (0.56)^8 = 0.00967
- Let's do part c before part b. For this, we want an exact booking, which means that exactly 7 of the 8 non-regular customers arrive for the flight. Suppose we align these 8 people in a row. Take the scenario that the 1st person didn't arrive and the remaining 7 did. That odds of that happening would be (1-.56)*(.56)^7.
Now take the scenario that the second person didn't arrive and the remaining 7 did. The odds would be
(0.56)(1-0.56)(0.56)^6 = (1-.56)*(.56)^7. You can run through every scenario that way and see that each time the odds are the same. There are a total of 8 different scenarios since we can choose 1 person (the non-arriver) from 8 people in eight different ways (combination).
So the overall probability of an exact booking would be [(1-.56)*(.56)^7] * 8 = 0.06079
- The probability that the flight has one or more empty seats is the same as the probability that the flight is NOT exactly booked NOR is it overbooked. Formally,
P(at least 1 empty seat) = 1 - P(-1 or 0 empty seats)
= 1 - P(overbooked) - P(exactly booked)
= 1 - 0.00967 - 0.06079
= 0.9295.
Note that, the chance of being both overbooked and exactly booked is zero, so we don't have to worry about that.
Hope that helps!
Have a great day :P
Answer:
yes if you give me brainliest
Step-by-step explanation:
