1.Since opening night, attendance at Play A has increased steadily, while attendance at Play B first rose and then fell. Equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. On what day(s) was the attendance the same at both plays? what was the attendance?
Play A: y=7x+84
Play B: y=-x^2+40x-6
A. The attendance was the same on day 30. The attendance was 294 at both place that day.****
B.The attendance was the same on day 3. The attendance was 105 at both place that day.
C. The attendance was never the same at both plays
D. The attendance was the same on days 3 and 30. The attendance at both plays on those days was 105 and 294 respectively.
I think the answer is A, please correct me if I'm wrong please.
well, A is true, but so is C. So, the answer is D.
If you had solved the quadratic you would have found both solutions
7x+84 = -x^2+40x-6
bro thats just wrong
The answer is actually d
Since opening night, attendance at Play A has increased steadily, while attendance at Play B first rose and then fell. Equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. On what day(s) was the attendance the same at both plays? What was the attendance?
Play A: y = 15x + 76
Play B: y = –x2 + 36x – 4
A. The attendance was the same on day 5. The attendance was 151 at both plays on that day.
B. The attendance was the same on day 16. The attendance was 316 at both plays on that day.
C. The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316 respectively.
D. The attendance was never the same at both plays.
The time t required to drive a certain distance varies inversely with the speed, r. If it takes 4 hours to drive the distance at 40 miles per hour, how long will it take to drive the same distance at 55 miles per hour?
A. about 2.91 hours
B. about 160.00 hours
C. about 22.00 hours
D. about 5.50 hours
The time and speed have an inverse relationship, which means that their product is constant. Let this constant be k. So, we can write:
t * r = k
When t = 4 hours and r = 40 miles per hour, we can find the value of k:
4 * 40 = k
k = 160
Now, we can use this value of k to find the time required to drive the same distance at 55 miles per hour:
t * 55 = 160
t = 160 / 55
t ≈ 2.91 hours (rounded to two decimal places)
Therefore, the answer is A. It will take about 2.91 hours to drive the same distance at 55 miles per hour.
The number of hours a group of contestants spends preparing for a quiz show are listed below. What is the frequency table that represents that data?
77 38 8 43 9 19 57 5 9 69 9 15 72 32 75 80 92
A. time chartZero to 9 hours has a frequency of 5. Ten to 19 hours has a frequency of 1. Twenty to 29 hours has a frequency of 1. Thirty to 39 hours has a frequency of 2. Forty to 49 hours has a frequency of 1. Fifty to 59 hours has a frequency of 1. Sixty to 69 hours has a frequency of 1. Seventy to 79 hours has a frequency of 3. Eighty to 89 hours has a frequency of 1. Ninety to 99 hours has a frequency of 1.
B. time chartZero to 9 hours has a frequency of 5. Ten to 19 hours has a frequency of 2. Twenty to 29 hours has a frequency of zero. Thirty to 39 hours has a frequency of 2. Forty to 49 hours has a frequency of 1. Fifty to 59 hours has a frequency of 1. Sixty to 69 hours has a frequency of 2. Seventy to 79 hours has a frequency of 2. Eighty to 89 hours has a frequency of 1. Ninety to 99 hours has a frequency of 1.
C. time chartZero to 9 hours has a frequency of 5. Ten to 19 hours has a frequency of 2. Twenty to 29 hours has a frequency of zero. Thirty to 39 hours has a frequency of 2. Forty to 49 hours has a frequency of 1. Fifty to 59 hours has a frequency of 1. Sixty to 69 hours has a frequency of 1. Seventy to 79 hours has a frequency of 3. Eighty to 89 hours has a frequency of 1. Ninety to 99 hours has a frequency of 1.
D. time chartZero to 9 hours has a frequency of 5. Ten to 19 hours has a frequency of 2. Twenty to 29 hours has a frequency of zero. Thirty to 39 hours has a frequency of 2. Forty to 49 hours has a frequency of 1. Fifty to 59 hours has a frequency of 1. Sixty to 69 hours has a frequency of 3. Seventy to 79 hours has a frequency of 1. Eighty to 89 hours has a frequency of zero. Ninety to 99 hours has a frequency of 2.
To create a frequency table, we need to count how many times each range of hours occurs in the data. We can create the following table:
Range of hours Frequency
0-9 5
10-19 2
20-29 0
30-39 2
40-49 1
50-59 1
60-69 1
70-79 3
80-89 0
90-99 1
Therefore, the answer is C. The frequency table that represents the data is: "Zero to 9 hours has a frequency of 5. Ten to 19 hours has a frequency of 2. Twenty to 29 hours has a frequency of zero. Thirty to 39 hours has a frequency of 2. Forty to 49 hours has a frequency of 1. Fifty to 59 hours has a frequency of 1. Sixty to 69 hours has a frequency of 1. Seventy to 79 hours has a frequency of 3. Eighty to 89 hours has a frequency of 0. Ninety to 99 hours has a frequency of 1."
What are the minimum, first quartile, median, third quartile, and maximum of the data set?
40, 7, 2, 35, 12, 23, 18, 28
A. minimum = 2
first quartile = 7
median = 20.5
third quartile = 35
maximum = 40
B. minimum = 2
first quartile = 7
median = 18
third quartile = 31.5
maximum = 40
C. minimum = 2
first quartile = 9.5
median = 23
third quartile = 35
maximum = 40
D. minimum = 2
first quartile = 9.5
median = 20.5
third quartile = 31.5
maximum = 40
22 / 32
To find the minimum, first quartile, median, third quartile, and maximum of the data set, we first need to put the data in order from smallest to largest:
2, 7, 12, 18, 23, 28, 35, 40
The minimum is the smallest value, which is 2.
To find the median, we need to find the middle value in the data set. Since there are 8 values, the median is the average of the fourth and fifth values:
median = (18 + 23) / 2 = 20.5
To find the first quartile, we need to find the median of the lower half of the data set. The lower half of the data set is:
2, 7, 12, 18
Since there are 4 values in the lower half, the first quartile is the average of the second and third values:
first quartile = (7 + 12) / 2 = 9.5
To find the third quartile, we need to find the median of the upper half of the data set. The upper half of the data set is:
23, 28, 35, 40
Since there are 4 values in the upper half, the third quartile is the average of the second and third values:
third quartile = (28 + 35) / 2 = 31.5
The maximum is the largest value, which is 40.
Therefore, the answer is D. The minimum is 2, the first quartile is 9.5, the median is 20.5, the third quartile is 31.5, and the maximum is 40.