146 area of trapazoid 90
area of triangle 56
Johnny is selling tickets to a school play. On the first day of ticket sales he sold 14 senior (S) citizen tickets and 4 child (C) tickets for a total of $200. On the second day of ticket sales he sold 7 senior (S) citizen tickets and 1 child (C) ticket for a total of $92. What is the price of one child ticket?
14S + 4C = 200
14S = 200 - 4C
S = (200 - 4C)/14
7S + 1C = 92
7S = 92 - C
S = (92 - C)/7
(200 - 4C)/14 = (92 - C)/7
7 x (200 - 4C) = 14 x (92 - C)
1400 - 28C = 1288 - 14C
1400 - 1288 = 28C - 14C
112 = 14C
C = 112/14 = 8
the price of one child ticket = $8
i am pretty sure it is 0.6
Answer:
Complete the following statements. In general, 50% of the values in a data set lie at or below the median. 75% of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 500 test scores, of them 0.5*500 = 250 would be at or below the median. If a sample consists of 500 test scores, of them 0.75*500 = 375 would be at or above the first quartile (Q1).
Step-by-step explanation:
The median separates the upper half from the lower half of a set. So 50% of the values in a data set lie at or below the median, and 50% lie at or above the median.
The first quartile(Q1) separates the lower 25% from the upper 75% of a set. So 25% of the values in a data set lie at or below the first quartile, and 75% of the values in a data set lie at or above the first quartile.
The third quartile(Q3) separates the lower 75% from the upper 25% of a set. So 75% of the values in a data set lie at or below the third quartile, and 25% of the values in a data set lie at or the third quartile.
The answer is:
Complete the following statements. In general, 50% of the values in a data set lie at or below the median. 75% of the values in a data set lie at or below the third quartile (Q3). If a sample consists of 500 test scores, of them 0.5*500 = 250 would be at or below the median. If a sample consists of 500 test scores, of them 0.75*500 = 375 would be at or above the first quartile (Q1).
THis equals the number of permutations of 4 from 6.
= 6P4 = 6! / (6 - 4)! = 6! / 2! = 720/2 = 360 answer
