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weeeeeb [17]
3 years ago
5

there was a total of $548 collected for the school play. the adult tickets cost $6 and the student tickets cost $4. if 12 more s

tudent tickets were sold than adult tickets, find the number of adult and student tickets sold.
Mathematics
2 answers:
balandron [24]3 years ago
7 0
You can first make the equations:

6x+4y=548
y-12=x

Here, you can use the substitution method and you get:

6(y-12)+4y=548
6y-72+4y=548
10y=620
y=62

Then there are 62 student tickets and 50 adult tickets.

Sonbull [250]3 years ago
3 0

Answer: There are 50 adult tickets and 62 student tickets.

Step-by-step explanation:

Since we have given that

Cost of adult ticket = $ 6

Cost of student ticket = $ 4

Total cost = $ 548

Number of adult ticket be 'x'.

Number of student ticket be 'x+12'

According to question, we get that,

6x+4(x+12)= 548

6x+4x+48 = 548

10x = 548-48

10 x = 500

x = 500 ÷ 10

x = 50

x + 12 = 50 + 12 = 62

Hence, there are 50 adult tickets and 62 student tickets.

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For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12
notsponge [240]
The binomial distribution is given by, 
P(X=x) =  (^{n}C_{x})p^{x} q^{n-x}
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n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) = (^{100}C_{x})(0.2)^{x} (0.8)^{100-x}
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∴ P(X≤12) = (^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1} + (^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3} + (^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5} + (^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7} + (^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9} + (^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11} + (^{100}C_{12})(0.2)^{12} (0.8)^{100-12}


Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability. 
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