Let me try all my best for this
Please , open the file
Answer:
about 0.177 mg/mL
Step-by-step explanation:
The maximum is found where the derivative of C(t) is zero.
dC/dt = 1.35e^(-2.802t) -(1.35t)2.802e^(-2.802t) = 0
Solving for t gives ...
t = 1/2.802
So, the maximum C(t) is ...
C(1/2.802) = 1.35/2.802e^(-1) ≈ 0.177 . . . . . mg/mL
The maximum average BAC during the first 6 hours is about 0.177 mg/mL.
_____
The maximum occurs about 21 minutes 25 seconds after consumption.
What is the upper quartile, Q3, of the following data set? 54, 53, 46, 60, 62, 70, 43, 67, 48, 65, 55, 38, 52, 56, 41
scZoUnD [109]
The original data set is
{<span>54, 53, 46, 60, 62, 70, 43, 67, 48, 65, 55, 38, 52, 56, 41}
Sort the data values from smallest to largest to get
</span><span>{38, 41, 43, 46, 48, 52, 53, 54, 55, 56, 60, 62, 65, 67, 70}
</span>
Now find the middle most value. This is the value in the 8th slot. The first 7 values are below the median. The 8th value is the median itself. The next 7 values are above the median.
The value in the 8th slot is 54, so this is the median
Divide the sorted data set into two lists. I'll call them L and U
L = {<span>38, 41, 43, 46, 48, 52, 53}
U = {</span><span>55, 56, 60, 62, 65, 67, 70}
they each have 7 items. The list L is the lower half of the sorted data and U is the upper half. The split happens at the original median (54).
Q3 will be equal to the median of the list U
The median of U = </span>{<span>55, 56, 60, 62, 65, 67, 70} is 62 since it's the middle most value.
Therefore, Q3 = 62
Answer: 62</span>
Answer:
6983/1000 OR 6.983
Step-by-step explanation:
6*1=6
9*1/10=9/10
8*1/100=8/100
3*1/1000=3/1000
6+9/10+8/100+3/1000=6983/1000
OR
6.983
Answer:
w=-7/2z
Step-by-step explanation:
to solve for w you firstly have to group the like terms
5w-3w=2z-9z
2w=-7z
then you have to divide both sides by 2 inorder to remain with w on one side.
2w/2=-7/2
w=-7/2
I hope this helps
