Answer: Ken Should have moved 9 units to the right because his prediction for May is 9°C. Moving to the left part of the number line would take Ken into negative degrees. 9°C is warmer than 7°C. Therefore, Ken should have moved 9 units to the right of the number line instead of 2 units to the left.
Step-by-step explanation:
Answer:
is this a yes or no question
(a) Data with the eight day's measurement.
Raw data: [60,58,64,64,68,50,57,82],
Sorted data: [50,57,58,60,64,64,68,82]
Sample size = 8 (even)
mean = 62.875
median = (60+64)/2 = 62
1st quartile = (57+58)/2 = 57.5
3rd quartile = (64+68)/2 = 66
IQR = 66 - 57.5 = 8.5
(b) Data without the eight day's measurement.
Raw data: [60,58,64,64,68,50,57]
Sorted data: [50,57,58,60,64,64,68]
Sample size = 7 (odd)
mean = 60.143
median = 60
1st quartile = 57
3rd quartile = 64
IQR = 64 -57 = 7
Answers:
1. The average is the same with or without the 8th day's data. FALSE
2. The median is the same with or without the 8th day's data. FALSE
3. The IQR decreases when the 8th day is included. FALSE
4. The IQR increases when the 8th day is included. TRUE
5. The median is higher when the 8th day is included. TRUE
Scale factor is equal to 1/2, since the point went from 10 to 5
Answer:
The possible values of the number of dollars in the original pile of money is ≥ $200 but < $350
Step-by-step explanation:
Here we have, pile of money ≥ $200
Amount in put the left pocket = $50
Fraction given away = 2/3 of rest of pile ≥ 2/3×150 ≥ $100
Amount put in right pocket = ≥ $150 - $100 ≥ $50
Total amount remaining with Jeri = $50 +≥ $50 ≥ $100
Also original pile - $200 < $100
Therefore where maximum amount given away to have more money = $200 we have
2/3× (original pile - 50) = $200
Maximum amount for original pile = $350
Therefore the possible values of the number of dollars in the original pile of money is ≥ $200 but < $350.
