This is an optimization calculus problem where you would need to know a little bit more about the box, atleast i would think. You would just need to use the volume equation of a sphere as the restrictive equation in the optimization problem. Perhaps there is a way to solve with the given information, but i do not know how to.
Range: maximum - minimum
IQR: third quartile - first quartile
median: middle number once all numbers are put in numerical order
Answer:
Step-by-step explanation:
The answer would be 11 gallons.
The tub starts with 32 gallons. Every minute it loses 3 gallons.
So after 1 minute it has lost 3 gallons. For example:
32-3= 29 Meaning that after 1 minute the tub has 29 gallons left.
Now you have to remember that the tub is draining for 7 minutes. So it is losing 3 gallons 7 times, because it loses 3 gallons each minute and there are 7 minutes.
We can use multiplication to find how many gallons the tub loses after 7 minutes. This sign “X” basically means “groups of”. We have 7 groups of 3, or
7 X 3
This is the same as saying we have 3, 7 times. Written like this:
3+3 +3 +3 +3 +3 +3 = 21
So after 7 minutes the tub has lost 21 gallons.
Now we take the original number of gallons and take away what was lost:
32-21=11
So there are 11 gallons left after 7 minutes.
Please let me know if you need any further explanation. Hope this helped.
Answer:
x = 18 y = 10
Step-by-step explanation:
let the first number be x
let the second number be y
x = y + 8..... equation 1
2x + y = 46.... equation 2
x is the larger number
y is the smaller number.
Rearrange the equation and add equation 1 to equation 2.
x - y = 8
+ 2x + y = 46
-------------------
3x + 0 = 54
3x = 54
divide both sides by 3
x = 54/3
x = 18
Substitute x = 18 into equation 1
x = y + 8
18 = y + 8
collect like terms
y = 18-8
y = 10
