Answer:
0.7061 = 70.61% probability she will have her first crash within the first 30 races she runs this season
Step-by-step explanation:
For each race, there are only two possible outcomes. Either the person has a crash, or the person does not. The probability of having a crash during a race is independent of whether there was a crash in any other race. This means that the binomial probability distribution is used 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.
A certain performer has an independent .04 probability of a crash in each race.
This means that 
a) What is the probability she will have her first crash within the first 30 races she runs this season
This is:

When 
We have that:



0.7061 = 70.61% probability she will have her first crash within the first 30 races she runs this season
Answer:
x ≈ 14.87
Step-by-step explanation:
Using Pythagoras' identity in the right triangle.
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is
x² = 10² + 11² = 100 + 121 = 221 ( take the square root of both sides )
x =
≈ 14.87 ( to the nearest hundredth )
Answer:
frequency of females watching action movie is 99
total participants 479
joint relative frequency of females and action movie = 99/479
your answer is D: divide 99 by 479
Step-by-step explanation:
Answer:
provide the value of B and C
The answer is 42.
The question is asking for angle N. We know from the question that angle N is equal to angle E (Don't think about the scale factor because it only applies to the sides. They're just trying to trick you).
To find angle E:

Move numbers to the right side:

Combine like terms:

Divide both sides by 13:

The formula for angle E:

Plug in the 8:

The angle measure of E is 42. So angle N is automatically 42 too.
To solidify, we can try to use the formula which they give us for angle N: