Number of child tickets bought is 20
<h3><u>
Solution:</u></h3>
Given that It cost 5 dollars for a child ticket and 8 dollars for a adult ticket
cost of each child ticket = 5 dollars
cost of each adult ticket = 8 dollars
Let "c" be the number of child tickets bought
Let "a" be the number of adult tickets bought
Total tickets sold were 110 bringing in 820 dollars
<em>Number of child tickets bought + number of adult tickets bought = 110</em>
c + a = 110 ----- eqn 1
<em><u>Also we can frame a equation as:</u></em>
Number of child tickets bought x cost of each child ticket + number of adult tickets bought x cost of each adult ticket = 820

5c + 8a = 820 -------- eqn 2
Let us solve eqn 1 and eqn 2 to find values of "c" and "a"
From eqn 1,
a = 110 - c ------ eqn 3
Substitute eqn 3 in eqn 2
5c + 8(110 - c) = 820
5c + 880 - 8c = 820
-3c = - 60
c = 20
Therefore from eqn 3,
a = 110 - 20 = 90
a = 90
Therefore number of child tickets bought is 20
Answer:
7
Step-by-step explanation:
9 is not a prime factor and of you divide 210 with 11, it will be a decimal
12,500 x 1.79 to find the price but 16,000 - 12,500 to get how many were left.
12,500 x 1.79 = $22,375 that they made from selling. 16,000 - 12,500 = 3,500 rolls left in the inventory. This question confused me so I just gave both answers.
Answer:
The Possible model is binomial distribution model.
Step-by-step explanation:
The argument that both students cheated in the exam can be proved by a hypothesis that both the students got the same answers incorrectly.
The same incorrect answers prove that both students have cheated on the test.
Therefore the sample of incorrect answers is, n = 8
Thus, the success probability, P = 0.25
Since the given condition has only two outcomes that are choosing the same answer or not choosing the same answer. Thus, this can be solved by the binomial distribution model.
So, binomial distribution with n = 8 and p = 0 .25.