Answer:
And we can find the individual probabilities using the probability mass function
And replacing we got:
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest "number of automobiles with both headligths working", on this case we now that:
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
And for this case we want to find this probability:
And we can find the individual probabilities using the probability mass function
And replacing we got:
Answer:
a = 85° , b = 30° , c = 65°
Step-by-step explanation:
∠ a and 85° are alternate angles and are congruent , so
a = 85°
∠ c and 65° are alternate angles and are congruent , so
c = 65°
a, b and c lie on a straight line and sum to 180° , that is
a + b + c = 180°
85° + b + 65° = 180°
150° + b = 180° ( subtract 150° from both sides )
b = 30°
Answer:
720
Step-by-step explanation:
1600*0.09=144
144*5=720
1)
where m is the slope
We know the slope is 5 and the line passes through the point (4,3).
The equation of the line is:
Now the second point:
The answer:
The value of y is -7.
2)
The line passes through the origin, or the point (0,0), and has a slope of -32.
The slope is negative so the line goes down from left to right.
It crosses both x and y axes in the point (0,0).
Therefore, it only passes through II and IV quadrant.
You can also find the equation of the line, find one more point and draw it.
Find the y-coordinate of the point for example (1,y).
The line passes through the points (0,0) and (1,-32). Now you can draw it (see the attachment).
The answer:
The line passes through quadrants II and IV.
