<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths.
Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday.
Jan - Friday
Feb - Monday
Mar - Monday
Apr - Thursday
May - Saturday
Jun - Tuesday
Jul - Thursday
Aug - Sunday
Sep - Wednesday
Oct - Friday
Nov - Monday
Dec - Wednesday
Now let's count how many times for each weekday, the 13th falls there.
Sunday - 1
Monday - 3
Tuesday - 1
Wednesday - 2
Thursday - 2
Friday - 2
Saturday - 1
The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc.
So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get:
Jan - Friday
Feb - Monday
Mar - Tuesday
Apr - Friday
May - Sunday
Jun - Wednesday
Jul - Friday
Aug - Monday
Sep - Thursday
Oct - Saturday
Nov - Tuesday
Dec - Thursday
And the weekday totals are:
Sunday - 1
Monday - 2
Tuesday - 2
Wednesday - 1
Thursday - 2
Friday - 3
Saturday - 1
And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year.
And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
Answer:
The probability that a part was manufactured on machine A is 0.3
Step-by-step explanation:
Consider the provided information.
It is given that Half of a set of parts are manufactured by machine A and half by machine B.
P(A)=0.5
Let d represents the probability that part is defective.
Ten percent of all the parts are defective.
P(d) = 0.10
Six percent of the parts manufactured on machine A are defective.
P(d|A)=0.06
Now we need to find the probability that a part was manufactured on machine A, and given that the part is defective
:
Hence, the probability that a part was manufactured on machine A is 0.3
Answer:
looks like A, first answer, is correct when using n such that it names what value of the sequence to use like a1 = 4 and so on.
Step-by-step explanation:
Given the table:
x f(x)
2 6
7 3
9 5
To determine the coordinate given in the table, we have the following:
f(2) = 6
f(7) = 3
f(9) = 5
From the answer choices, the only coordinate given is:
f(5) = 9
ANSWER:
f(9) = 5
