<u>Answer:</u>
The number of tickets bought is 1000 adults tickets and 700 children were present
<u>Explanation:</u>
Let the number of children and adults present in the circus be x and y respectively
According to question,
Then, x + y = 1700 ……………..equation (1)
And, cost of ticket to circus for children and adult be 19, 42 respectively
Total cost of ticket;
for adults = 42y; for children = 19x
19x + 42y = 55300 …………………equation (2)
Multiplying equation (1) by 19
19x + 19y = 32300 ……………… eqn (3)
Subtracting equation 1 and 3, we get
23y = 23000
y = 1000
putting this value in eq (1), we get x = 700
Therefore, 1000 adults and 700 children were present
.
Which data set has an outlier? 25, 36, 44, 51, 62, 77 3, 3, 3, 7, 9, 9, 10, 14 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 63, 65, 66,
umka21 [38]
It's hard to tell where one set ends and the next starts. I think it's
A. 25, 36, 44, 51, 62, 77
B. 3, 3, 3, 7, 9, 9, 10, 14
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Let's go through them.
A. 25, 36, 44, 51, 62, 77
That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.
B. 3, 3, 3, 7, 9, 9, 10, 14
Average around 7, sigma around 4, within 2 sigma, seems ok.
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
Average around 20, sigma around 8, that 39 is hanging out there past two sigma. Let's reserve judgement and compare to the next one.
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Average around 74, sigma 8, seems very tight.
I guess we conclude C has the outlier 39. That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.
Answer:
x = 5
x = 0.5
Step-by-step explanation:
2x2 - 11x + 5 = 0
Roots: 5, 0.5
Root Pair: 11/4 ± 9/4
Factored: f(x) = 2(x - 5)(x - 0.5)
