The equation of a circle with center and radius
is
In your case,
Plug these values into the generic formula and you'll get your equation.
Course Hero The Pretty Pizza Company wants to open a new pizza place on either the north side of town or the south side of town.
1 answer:
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Answer:
a) There is a 1.21% probability that both contain diet soda.
b) There is a 79.21% probability that both contain diet soda.
c) is unusual,
is not unusual
d) There is a 19.58% probability that exactly one is diet and exactly one is regular.
Step-by-step explanation:
There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.
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 combinatios of x objects from a set of n elements, given by the following formula.
And is the probability of X happening.
A number of sucesses is considered unusually low if
and unusually high if
In this problem, we have that:
Two cans are randomly chosen, so
Two out of 18 cans are filled with diet coke, so
a) Determine the probability that both contain diet soda. P(both diet soda)
That is P(X = 2).
There is a 1.21% probability that both contain diet soda.
b)Determine the probability that both contain regular soda. P(both regular)
That is P(X = 0).
There is a 79.21% probability that both contain diet soda.
c) Would this be unusual?
We have that is unusual, since
For P(X = 0), it is not unusually high nor unusually low.
d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)
That is P(X = 1).
There is a 19.58% probability that exactly one is diet and exactly one is regular.
Polynomials in the fourth degree are called quartic equations. In solving the roots of polynomials, there are techniques available. For quadratic equations, you use the quadratic formula. For cubic equations, you use the scientific calculator. But for quartic equations and higher, it is very complex. The method is very lengthy and can get very messy because you introduce a lot variables. So, I suggest you do the easiest method to estimate the roots.
Graph the equation by plotting arbitrary points. The graph looks like that in the figure. The points at which the curve passes the x-axis are the solution which are encircled in red.In approximation, the rational roots or zero's are -3.73, -1, -0.28 and 2.
Graph the equation by plotting arbitrary points. The graph looks like that in the figure. The points at which the curve passes the x-axis are the solution which are encircled in red.In approximation, the rational roots or zero's are -3.73, -1, -0.28 and 2.