Answer:
Step-by-step explanation:
information from the range
If the range is 26 and the largest number is 31 then the smallest number is 31 - 26 = 5
Information from the median
The median must consist of 2 numbers because we have 6 integers for the data and median must consist of the middle 2. Since we are told that the median is 15 and we need 2 integers the other integer must be 15 as well..
So what do we know so far.
5 15 15 31
Information from the mean
There are six integers making up the series of numbers. The mean is 16. Therefore all 6 integers added together must be 6 * 16 = 96
What do we have so for?
5 + 15 + 15 + 31 = 66
What do the other two numbers sum to?
96 -66 = 30
Getting the answer.
The first number to try must be 13. It has to be below 15 (the median and odd. That's 13).
So now we have
5 13 15 15 ___ 31
But from the mean we need 30
30 - 13 = 17
And we have a winner because 17 is above 15.
Answer
5 13 15 15 17 31
Answer:
Area of shaded region=41 square feet
Area of non shaded region= 87 square feet
Step-by-step explanation:
As, Shaded area is made of different shapes including two rectangles and one triangle
So,
Area of shaded region,

Area of non shaded region=total area of rectangular region-Shaded region

Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min