Add 8 on both sides in order to isolate n:
-3 + 8 < n - 8 + 8
5 < n
A single die is rolled twice. the set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),
Maksim231197 [3]
<span>We need to find the rolls whose sum is greater than 10. By looking at the outcomes, we see that (5,6), (6,5), and (6,6) all have a sum greater than 10. Therefore, there are 3 chances to get a sum greater than 10. Since there are 36 chances overall, the probability of rolling greater than 10 are 3/36 = 1/12.</span>
The radius of the circle expressed as a mixed number is: 
inches.
<h3>What is the Radius of a Circle?</h3>
Radius of a circle = half the measure of the diameter of a circle.
Given:
diameter of a circle = 
inches
Therefore:
Radius of the circle = 1/2()
= 1/2(47/6)
= (1 × 47)/(2 × 6)
= 47/12
= inches.
Therefore, the radius of the circle expressed as a mixed number is: inches.
Learn more about radius of a circle on:
brainly.com/question/24375372
Answer:
19.5
Step-by-step explanation:
Here is a key factors in triangles, it must add up to 180 degrees.
You have three angles in a triangle ABC,
A = 127
B = 33.5
C = ?
In this case you would add the two together (160.5) and subtract from 180 giving you the angle C = 19.5
CHECK:
Add 127+33.5+19.5 = 180! Yes!
PLs mark brainliest if you can! :)
<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>
