Every weekday, Tasha walks 3/8 mile to go to school. After school, she walks 2/8 mile to go to her grandparents' house. Then she takes the smae route home. Does she walk more than one mile each day? Explain how you know.
Answers
There are two possibilities, and we don't know which situation is true.
#1). School is in the middle, between home and grandparents.
Home --------- 3/8 ----------- School --------- 2/8 ------- Grandparents .
If this is true, then in the whole day, Tasha walks
(3/8 + 2/8 + 2/8 + 3/8) = 10/8 = 1.25 miles .
===========================================
#2): Grandparents house is in the middle, between home and school.
Tasha passes by them on her way to school, and stops there to visit
on her way back home.
Home ------- 1/8 ------- Grandparents ---------- 2/8 ---------- School
If this is true, then in the whole day, Tasha walks
(3/8 + 2/8 + 1/8) = 6/8 = 3/4 mile .
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X is 6 more than twice y.
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(x,y)
we have
(-3,-3)
(7,-3)
(7,2)
so list known ones
x=-3,7,7
y=-3,-3,2
we find pairs
x=7 and 7
y=-3 and -3
the missing one is another -3 for x and 2 for y
therefor the last point is
(-3,2)
Answers
Answer:
7 is the median
Step-by-step explanation:
listing the numbers from smallest to largest
3,4,7,8,9
the median is the middle number
Answers
Final answer:
To convert feet to inches, multiply by 12. So, 3 feet is 36 inches. Therefore, the ratio of 28 inches to 3 feet is 28:36, which simplifies to 7:9
Explanation:
The ratio of 28 inches to 3 feet is usually used when comparing these two units of measurement. Since 1 foot is equivalent to 12 inches, then 3 feet is 36 inches.
Therefore, the ratio of 28 inches to 3 feet, expressed in inches, would be 28:36 which can be simplified to 7:9 by dividing both numbers by 4.
The ratio shows a direct relationship between the two units, meaning that for every 7 units on the '28 inches' side, there are 9 units on the '3 feet' side.
Learn more about Ratio Conversion here:
#SPJ2
The total cost of buying 20 cakes is £T.
Write a formula for T in terms of C.
thanks
:):):):):):):)::)::):):::)):):):):):):):):):):):):):):):):)):):):):):)):):):):):):):):
Answers
and 20 cakes is T
So 1 cake times 20=20 cakes
so 20C=T
Answers
the median of the temperatures at Meadows: ______
the interquartile range of the temperatures at Springwood:_______
the interquartile range of the temperatures at Meadows:_______
the difference of the medians as a multiple of their average interquartile range:_______
Answers
Hello,
The median of the temperatures at Springwood is 86. (median is at the middle).
the median of the temperatures at Meadows is 73. (median is at the middle).
The (IQR) interquartile range of the temperatures at Meadows is 12. (You have to subtract both of the IQR left and right 80 - 68 = 12).
The (IQR) interquartile range of the temperatures at Springwood is 14. (You have to subtract both of the IQR left and right 91 - 77 = 14).
the difference of the medians as a multiple of their average interquartile range is 79.5 is average median and 13 average IQR.
Answer: The median of the temperatures at Springwood is 86.
The median of the temperatures at Meadows is 73.
The interquartile range of the temperatures at Meadows is 12.
The interquartile range of the temperatures at Springwood is 14.
The difference of the medians as a multiple of their average interquartile range as : Difference in medians = 13= 1 x 13
i.e. Difference in medians = 1 x (Average interquartile range)
Step-by-step explanation:
In a box - whisker plot , the vertical line in box represents the median value.
The left end of box denotes Lower quartile and the right end of the box Upper quartile.
By considering the given picture,
For Meadows ,
Median = 73
The median of the temperatures at Meadows is 73. (1)
Lower quartile = 68
Upper quartile = 80
Interquartile range = Upper quartile- Lower quartile
Interquartile range=80-68 = 12
The interquartile range of the temperatures at Meadows is 12. (2)
For Springwood,
Median = 86
The median of the temperatures at Springwood is 86. (3)
Lower quartile = 77
Upper quartile = 91
Interquartile range = Upper quartile- Lower quartile
= 91-77 = 14
The interquartile range of the temperatures at Springwood is 14. (4)
Difference in medians = 86-73 = 13 [From (1) and (3)]
Average interquartile range = [From (2) and (4)]
The difference of the medians as a multiple of their average interquartile range as : Difference in medians = 13= 1 x 13
i.e. Difference in medians = 1 x (Average interquartile range)