In a school of 450 people 110 are in the choir 240 are in a band and 60 are in bothWhat is the probability
Answers
The question doesn't require any specific probabilities, but I'm adding my own calculations to make it easier for you to solve your own problem
Answer:
Questions added and answered below
Step-by-step explanation:
Venn Diagram
When we have different sets, some of them belong only to one set, some belong to more than one, some don't belong to any of them. This situation can be graphically represented by the Venn Diagrams.
Let's analyze the data presented in the problem and fill up the numbers into our Venn Diagram. First, we must use the most relevant data: there are 60 people in both the choir and the band. This number must be in the common space between both sets in the diagram (center zone, purple).
We know there are 110 people in the choir, 60 of which were already placed in the intersection zone, so we must place 110-60=50 people into the blue zone, belonging to C but not to B.
We are also told that 240 people are in a band, 60 of which were already placed in the intersection zone, so we must place 240-60=180 people into the red zone, belonging to B but not to C.
Finally, we add the elements in all three zones to get all the people who are in the choir or in the band, and we get 50+60+180=290. Since we have 450 people in the school, there are 450-290=160 people who are not in the choir nor in the band.
The question doesn't ask for a particular probability, so I'm filling up that gap with some interesting probability calculations like
a) What is the probability of selecting at random one person who is in the band but not in the choir?
The answer is calculated as
b) What is the probability of randomly selecting one person who belongs only to one group?
We look for people who are in only one of the sets, they are 50+180=230 people, so the probability is
b) What is the probability of selecting at random one person who doesn't belong to the choir?
We must add the number of people outside of the set C, that is 180+160=340
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Can someone help in this one please
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P - Parenthesis
E - Exponents
M - Multiplication
D - Division
A - Addition
S - Subtraction
Then the special cases:
(I have attached a picture below)
Lets get into action!
(5a - 2)(5a + 2) - 3( 1 - a ) - 3a
25a² - 4 - 3 + 3a - 3a
25a² - 7
25a² - 7 is the answer
(5a-2)(5a+2)-3(1-a)-3a=25a2−7
Therefore answer is 25a2−7
Answers
75/1 x 8/1 = 600/1
75 x 8 = 600
Answers
Please answer these two math questions for me :)
Answers
"Please tell Kat that Brainly gives half the points she selects to each user that answers the question."
The Answers
#1:
We are trying to find the MAD for each set of the teachers' grades. To find the MAD (again) find the mean of the teachers' grades. Then, take each of the teachers' grades and find the absolute value of the difference between that grade and the mean grade. Add the differences up, and divide the result by the number of teachers (10).
Do the first table first (first things first!). Find the mean by adding up the grades and dividing them by the number of teachers (10).
76 + 81 + 85 + 79 + 89 + 86 + 84 + 80 + 88 + 79 = 827
827/10 = 82.7
The mean grade for the first table is 82.7.
Now, we need to find the absolute value of the difference between each of the teachers' grades and the mean.
82.7 - 76 = 6.7
82.7 - 81 = 1.7
85 - 82.7 = 2.3
82.7 - 79 = 3.7
89 - 82.7 = 6.3
86 - 82.7 = 3.3
84 - 82.7 = 1.3
82.7 - 80 = 2.7
88 - 82.7 = 5.3
82.7 - 79 = 3.7
Now, we need to add up the result.
6.7 + 1.7 + 2.3 + 3.7 + 6.3 + 3.3 + 1.3 + 2.7 + 5.3 + 3.7 = 37
Now, we divide the result by the number of teachers.
37/10 = 3.7
So the MAD for the first table is 3.7.
Now we do the same thing for the second table.
79 + 82 + 84 + 81 + 77 + 85 + 82 + 80 + 78 + 83 = 811
811/10 = 81.1
Bad grades.
81.1 - 79 = 2.1
82 - 81.1 = 0.9
84 - 81.1 = 2.9
81.1 - 81 = 0.1
81.1 - 77 = 4.1
85 - 81.1 = 3.9
82 - 81.1 = 0.9
81.1 - 80 = 1.1
81.1 - 78 = 3.1
83 - 81.1 = 1.9
Add.
2.1 + 0.9 + 2.9 + 0.1 + 4.1 + 3.9 + 0.9 + 1.1 + 3.1 + 1.9 = 21
21/10=2.1
Now, we compare the MAD's. If the MAD for the second table is greater than the MAD for the first table, then the teachers have not improved/were sleeping. If not, then the teachers have improved/got lucky.
Hooray! They improved/got lucky, by 1.6 points to be precise.
#2:
Remember from the previous answer I gave (the other question you asked) the example with a MAD of 0? That's because they were all exactly on the mean (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 40, 40/10 = 4, MAD = 0). Therefore, if the MAD of the teachers' grades was zero, that means they all got the same score! Hooray! Mission accomplished. Or the common core machine has succeeded in producing identical copies of teachers.
Backword
I hope you know what the MAD is now.
Answers
Need 2 ways!!! 2 answers!!!!!!!!! Please! ASAP!
Answers
Mandy sold: 8,900 shirts in a week.
Key word: Less
We need to subtract.
8900 - 765 = 8,135
Rae sold 8,135 shirts LESS than Mandy.
OR Mandy sold 8,135 MORE shirts than Rae.
So Rae sold 8135 LESS shirts and....
Mandy sold 8135 MORE shirts!