In a symmetric histogram, you have the same number of points to the left and to the right of the median. An example of this is the distribution {1,2,3,4,5}. We have 3 as the median and there are two items below the median (1,2) and two items above the median (4,5).
If we place another number into this distribution, say the number 5, then we have {1,2,3,4,5,5} and we no longer have symmetry. We can fix this by adding in 1 to get {1,1,2,3,4,5,5} and now we have symmetry again. Think of it like having a weight scale. If you add a coin on one side, then you have to add the same weight to the other side to keep balance.
Answer A is the only one that seems to make sense to me id choose A
A proportional relationship is a relationship between variables such that the amount of one variable is a multiple of the other variable
A table showing the amount of sugar in different sizes of orange juice is presented as follows;
![\begin{array}{|c|cc|}\mathbf{Amount \ of \ orange \ juice, \ x}&&\mathbf{Quantity \ of \, sugar\ (g), \ f(x)}\\3&&8.25\\6&&16.5\\9&&24.75\\12&&33\\15&&41.25\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Ccc%7C%7D%5Cmathbf%7BAmount%20%5C%20of%20%5C%20orange%20%5C%20juice%2C%20%5C%20x%7D%26%26%5Cmathbf%7BQuantity%20%5C%20of%20%5C%2C%20sugar%5C%20%28g%29%2C%20%5C%20f%28x%29%7D%5C%5C3%26%268.25%5C%5C6%26%2616.5%5C%5C9%26%2624.75%5C%5C12%26%2633%5C%5C15%26%2641.25%5Cend%7Barray%7D%5Cright%5D)
Reason:
The given parameter are;
12 ounces of orange juice contains 33 grams of sugar
Therefore, we have;
The amount of sugar per ounce of orange juice is given as follows;

Therefore, there are 2.75 grams of sugar per ounce of orange juice
which gives;
The amount of sugar in different sizes of orange juice, f(x) = 2.75·x
Where <em>x</em> represents the amount of orange juice, and f(x) represents the amount of sugar in the orange juice
A table showing the amount of sugar present in different sizes of orange juice is given as follows;
![\begin{array}{|c|cc|}\mathbf{Amount \ of \ orange \ juice, \ x}&&\mathbf{Quantity \ of \, sugar\ (g), \ f(x)}\\3&&8.25\\6&&16.5\\9&&24.75\\12&&33\\15&&41.25\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Ccc%7C%7D%5Cmathbf%7BAmount%20%5C%20of%20%5C%20orange%20%5C%20juice%2C%20%5C%20x%7D%26%26%5Cmathbf%7BQuantity%20%5C%20of%20%5C%2C%20sugar%5C%20%28g%29%2C%20%5C%20f%28x%29%7D%5C%5C3%26%268.25%5C%5C6%26%2616.5%5C%5C9%26%2624.75%5C%5C12%26%2633%5C%5C15%26%2641.25%5Cend%7Barray%7D%5Cright%5D)
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there are 4 9's in a deck so probability would be 4/52 reduced to 1/13
does the answer need to be in a different format?