Solving:
when you divide, you get this number: 1.56775
You have been asked to round to the tenth.
The tenth is one place right of the decimal. (currently a 5)
Look at the next place to the right. Since it is 6, we round up the tenth place by one. So our 5 in the tenth place becomes 6 (because the next digit was greater than 4)
1.6 is the final result
The answer is the mean, mode, and median increases by 4, the range of times is the same.
Week 1: Week 2:
Student - Hours Student - Hours<span>
Bob 19 </span>Bob 23<span>
James 10 </span>James 14<span>
Karen 15 </span>Karen 19<span>
Rosario 17 </span>Rosario 21<span>
Antoine 10 </span>Antoine 14<span>
Julio 16 </span>Julio 20<span>
Maria 13 </span>Maria 17<span>
The mean is the sum of all values divided by the number of values:
Week 1: (19 + 10 + 15 + 17 + 10 + 16 + 13)/7 = 100/7 = 14.28
Week 2: (23 + 14 + 19 + 21 + 14 + 20 + 17)/7 = 128/7 = 18.28
The difference in means between Week 2 and Week 1 is 4 (18.28 - 14.28 = 4)
The median is the middle value. To calculate, first rearrange values from the lowest to the highest and then find the middle value:
Week 1: 10, 10, 13, 15, 16, 17, 19 - The median is 15.
Week 2: 14, 14, 17, 19, 20, 21, 23 - The median is 19.
The difference in medians between Week 2 and Week 1 is 4 (19 - 15 = 4)
The mode is the value that occurs most frequently.
</span>Week 1: 10, 10, 13, 15, 16, 17, 19 - The mode is 10.
Week 2: 14, 14, 17, 19, 20, 21, 23 - The mode is 14.
The difference in modes between Week 2 and Week 1 is 4 (14 - 10 = 4)
The range of times is the difference between the highest and the lowest value.
Week 1: 10, 10, 13, 15, 16, 17, 19 - The range of times is 9 (19 - 10 = 9).
Week 2: 14, 14, 17, 19, 20, 21, 23 - The median is 9 (23 - 14 = 9).
The difference in the ranges of times between Week 2 and Week 1 is 0 (9 - 9 = 0)
Answer:
32
Step-by-step explanation:
Because if you put the lack of other side to the other side, it will complete.
One roll = 4
eight rolls = 4 × 8 = 32
If f is a one-to-one continuous function defined on an interval, then its inverse f−1 is also one-to-one and continuous.
Answer: C Its a non-perfect Square.
