The ratio of 7200 to 6400 equals 1.125 in decimal form. I hope this helps!:)
Adding (or subtracting) a constant to every data value adds (or subtracts) the same constant to measures of position such as center,percentiles, max or min.
Its shape and spread such as range, IQR, standard deviation remain unchanged.
When we multiply (or divide) all the data values by any constant, all measures of position (such as the mean, median, and percentiles) and measures of spread (such as the range, the IQR, and the standard deviation) are multiplied (or divided) by that same constant.
Part A:
The lowest score is a measure of location, so both addition and multiplying the lowest score of test B by 40 and adding 50 to the result will affect the lowest score of test A.
Thus, the lowest score of test A is given by 40(21) + 50 = 890
Therefore, the lowest score of test A is 890.
Part B:
The mean score is a measure of location, so both
addition and multiplying the mean score of test B by 40 and adding 50
to the result will affect the lowest score of test A.
Thus, the mean score of test A is given by 40(29) + 50 = 1,210
Therefore, the mean score of test A is 890.
Part C:
The standard deviation is a measure of spread, so multiplying the standard deviation of test B by 40 will affect the standard deviation but adding 50
to the result will not affect the standard deviation of test A.
Thus, the standard deviation of test A is given by 40(2) = 80
Therefore, the standard deviation of test A is 80.
Part D
The Q3 score is a measure of location, so both
addition and multiplying the Q3 score of test B by 40 and adding 50
to the result will affect the Q3 score of test A.
Thus, the Q3 score of test A is given by 40(28) + 50 = 1,170
Therefore, the Q3 score of test a is 1,170.
Part E:
The median score is a measure of location, so both
addition and multiplying the median score of test B by 40 and adding 50
to the result will affect the median score of test A.
Thus, the median score of test A is given by 40(26) + 50 = 1,090
Therefore, the median score of test A is 1,090.
Part F:
The IQR is a measure of spread, so multiplying the IQR of test B by 40 will affect the IQR but adding 50
to the result will not affect the IQR of test A.
Thus, the IQR of test A is given by 40(6) = 240
Therefore, the IQR of test A is 240.
Answer:
100 students on each bus
Step-by-step explanation:
100*4 is 400 so there might be just 4
2x-10=x+15
X-10=15
X=25
Bc the of angle 5 is congruent to angle 7
Answer:
1:30pm
Step-by-step explanation:
Start Time: 10:00am
Start Speed: 40MPH
Time for Lunch: 30minutes
Arrival Time: 5:00pm
Ending Speed: 50MPH
Total Miles: 290miles
Find Total Time Taken:
5pm - 10am = 7 hours
7 hours - 30 minutes (lunch) = 6.5 hours of travel
Create Equations:
Distance = Speed x Time
Finding Total Numbers of Miles Traveled
Miles = SpeedBefore x TimeBefore + SpeedAfter x TimeAfter
290(Miles) = 40(mph) x E + 50(mph) x Y
290 = 40E + 50A
Finding Time Traveled Before and After the lunch break
Note: It is based on duration in hours
Total Time = Time Before + Time After + Lunch Time
7 hours = E + Y + .5 hours
7 = E + Y + .5
Solve for Y
7 = Y + E + .5
6.5 = Y + E
6.5 - E = Y
Use Substitution in second equation to solve for E
290 = 40E + 50Y
290 = 40E + 50(6.5 - E)
290 = 40E + (325 - 50E)
290 = 40E + 325 - 50E
290 = -10E + 325
-35 = -10E
10E = 35
E = 3.5 hours, time traveled before lunch
Note: The following is unnecessary
Solve for Y, Again
Y = 6.5 - E
Y = 6.5 - 3.5
Y = 3 hours, time traveled after lunch
So the man traveled for 3.5 hours and then ate lunch. Then he traveled for 3 more hours.
So...
10am + 3.5 hours = 1:30pm - the time when he ate lunch
