<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>
So she earned 84 total dollars
she earned 14 dollars mowing lawns
she earned (total-mowing lawns=babysitting) 84-14=70 dollar babysitting
7 dollars per hour so
70/7=10 so
h=10
Answer:
3 treats each.
Step-by-step explanation:
3 cats divided by 10 = 9 remainder 1 and 3 times 3 = 9.
Answer:
P(X > 10), n = 15, p = 0.7
P(X > 10) =P(10 < X ≤ 15) = P(11 ≤ X ≤ 15) = P(X = 11, 12, 13, 14, 15)
=P(X = 11) + P(X =12) + P(X = 13) + P(X =14) + P(X = 15) (because these are disjoint events)
Step-by-step explanation:
See attached image for detailed explanation
Answer:
he grows by 5 cm every year between 1999 and 2006
Step-by-step explanation:
This is a arithmetic progression problem with the formula;
T_n = a + (n - 1)d
We are told that In 1999 Daniel was 146 cm tall. He grew to be 176 cm by the year 2006.
Thus;
a = 146
d = 2006 - 1999 = 7
Thus;
176 = 146 + (7 - 1)d
176 - 146 = 6d
30 = 6d
d = 30/6
d = 5 cm
Thus, he grows by 5 cm every year between 1999 and 2006
