Answer:
(b) 1/792
Step-by-step explanation:
The complete question is;
<em>Counting: Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade.</em>
<em>(a) How many different groups of 5 problems can be chosen from the 12</em>
<em>problems?</em>
<em>(b)Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?</em>
<u>In (a)</u>
<u></u>
Apply the formula
where n=12 and r=5
substitute values
In (b)
If Jerry did only 5 problems of one assignment then the probability will be
<em />
<em />
Answer:
Step-by-step explanation:
The origin (x,y) is (0,0)
If its moved four units right you must add 4 to the x coordinate giving you (4,0)
Then it is moved up by 5, so you must add 5 to the y coordinate giving you
(4,5)
Answer:
1-Arc PQ is congruent to arc SR.
2-The measure of arc QR is 150°.
4-Arc PS measures about 13.1 cm.
5-Arc QS measures about 15.7 cm
Step-by-step explanation:
1,2,4,5
1-9 = 9 digits
10-99 = 180 digits
So if we continue the pattern to 99, there are 189 digits, and the last 5 digits would be 79899. Counting backwards: 189th = 9, 188th = 9, 187th = 8, 186th = 9, 185th = 7.
The 185th digit is 7.
<h3>
Answer:</h3>
equations
solution
<h3>
Step-by-step explanation:</h3>
Let "a" and "c" represent the numbers of adult and children's tickets sold, respectively. The problem statement tells us two relationships between these values:
... 20a +10c = 15000 . . . . . . total revenue from ticket sales
... c = 3a . . . . . . . . . . . . . . . . relationship between numbers of tickets sold
Using the expression for c, we can substitute into the first equation to get ...
... 20a +10(3a) = 15000
... 50a = 15000
... a = 15000/50 = 300 . . . . . adult tickets sold
... c = 3·300 = 900 . . . . . children's tickets sold
