Lowest wind chill temperature on Monday was 1/4(-36) = -9 F
The lakes initial level with respect of normal is -8 inches
New level after decreases = -8 - 1 1/4 - 2 3/8 = -11 5/8 inches.
I believe the answer is D.
Based on the capacity of the container and the capacity currently being used, the amount of water in the container is <u>3/8 gallons. </u>
The container is capable of holding 1/2 gallons of water.
If it is holding 3/4 of that 1/2 gallons, the amount of water it would be holding is:
<em>= Capacity of container x Percentage full </em>
= 1/2 x 3/4
= 3/8 gallons of water
In conclusion, the container is currently holding 3/8 gallons of water.
<em>Find out more about this at brainly.com/question/11459708. </em>
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given set of values

STEP 2: Write the formula for calculating the Standard deviation of a set of numbers
![\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ where\text{ }x_i\text{ are data points,} \\ \bar{x}\text{ is the mean} \\ \text{n is the number of values in the data set} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20S%5Ctan%20dard%5Ctext%7B%20deviation%3D%7D%5Csqrt%5B%5D%7B%5Cfrac%7B%5Csum%5E%7B%7D_%7B%7D%28x_i-%5Cbar%7Bx%7D%29%5E2%7D%7Bn-1%7D%7D%20%5C%5C%20where%5Ctext%7B%20%7Dx_i%5Ctext%7B%20are%20data%20points%2C%7D%20%5C%5C%20%5Cbar%7Bx%7D%5Ctext%7B%20is%20the%20mean%7D%20%5C%5C%20%5Ctext%7Bn%20is%20the%20number%20of%20values%20in%20the%20data%20set%7D%20%5Cend%7Bgathered%7D)
STEP 3: Calculate the mean

STEP 4: Calculate the Standard deviation
![\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ \sum ^{}_{}(x_i-\bar{x})^2\Rightarrow\text{Sum of squares of differences} \\ \Rightarrow10332.7225+657.9225+18591.3225+982.8225+2740.52251+9731.8225+3522.4225+18319.6225+2878.3225 \\ +8163.1225+1417.5225+3925.0225+1321.3225+386.1225+5677.6225+2953.9225+3800.7225 \\ +3209.2225+2565.4225+10537.0225 \\ \text{Sum}\Rightarrow108974.0275 \\ \\ S\tan dard\text{ deviation}=\sqrt[]{\frac{111714.55}{20-1}}=\sqrt[]{\frac{111714.55}{19}} \\ \Rightarrow\sqrt[]{5879.713158}=76.67928767 \\ \\ S\tan dard\text{ deviation}\approx76.68 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20S%5Ctan%20dard%5Ctext%7B%20deviation%3D%7D%5Csqrt%5B%5D%7B%5Cfrac%7B%5Csum%5E%7B%7D_%7B%7D%28x_i-%5Cbar%7Bx%7D%29%5E2%7D%7Bn-1%7D%7D%20%5C%5C%20%5Csum%20%5E%7B%7D_%7B%7D%28x_i-%5Cbar%7Bx%7D%29%5E2%5CRightarrow%5Ctext%7BSum%20of%20squares%20of%20differences%7D%20%5C%5C%20%5CRightarrow10332.7225%2B657.9225%2B18591.3225%2B982.8225%2B2740.52251%2B9731.8225%2B3522.4225%2B18319.6225%2B2878.3225%20%5C%5C%20%2B8163.1225%2B1417.5225%2B3925.0225%2B1321.3225%2B386.1225%2B5677.6225%2B2953.9225%2B3800.7225%20%5C%5C%20%2B3209.2225%2B2565.4225%2B10537.0225%20%5C%5C%20%5Ctext%7BSum%7D%5CRightarrow108974.0275%20%5C%5C%20%20%5C%5C%20S%5Ctan%20dard%5Ctext%7B%20deviation%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B111714.55%7D%7B20-1%7D%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B111714.55%7D%7B19%7D%7D%20%5C%5C%20%5CRightarrow%5Csqrt%5B%5D%7B5879.713158%7D%3D76.67928767%20%5C%5C%20%20%5C%5C%20S%5Ctan%20dard%5Ctext%7B%20deviation%7D%5Capprox76.68%20%5Cend%7Bgathered%7D)
Hence, the standard deviation of the given set of numbers is approximately 76.68 to 2 decimal places.
STEP 5: Calculate the First and third quartile

STEP 6: Find the Interquartile Range

Hence, the interquartile range of the data is 116
Just by adding both of these, pens and also pencils, these actually both signify the same word, "pencils".
We do . . . .

(
93) <em>should </em>be your correct answer!