Answer: 330 yd³
Step-by-step explanation:
v = whl
v = 6 x 11 x 5
v = 66 x 5
v = 330 yd³
Answer:
(D) Quantitative
Step-by-step explanation:
Bivariate data involves two types of variables. Categorical data will have a number of different categories. Continuous data usually includes a range of values so is likely to be in decimals. Quantitative data will be collected in this scenario as it will be calculated the number of times each shirt is tried on.
If

is odd, then

while if

is even, then the sum would be

The latter case is easier to solve:

which means

.
In the odd case, instead of considering the above equation we can consider the partial sums. If

is odd, then the sum of the even integers between 1 and

would be

Now consider the partial sum up to the second-to-last term,

Subtracting this from the previous partial sum, we have

We're given that the sums must add to

, which means


But taking the differences now yields

and there is only one

for which

; namely,

. However, the sum of the even integers between 1 and 5 is

, whereas

. So there are no solutions to this over the odd integers.
Answer:
-10
Step-by-step explanation:
189
Because you have to turn it into a percentage in order to know how many per 1 percent. So multiply 420 by 100 and get 4.2 then you multiply it by 45 and you get 189 so 179 schools have a horse mascot