All categories

3 years ago

A rectangular prism has a length of 5 1/8 feet, a width of 7 1/2 feet, and a height of 2 feet.

Mathematics

1 answer:

kipiarov [429]3 years ago

8 0

Answer:

If I'm correct than the answer should be 76.88 ft cubed

Step-by-step explanation:

5 1/8 would be 5.125

7 1/2 would be 7.5

Multiply 5.125 and 7.5 to get 38.44

Then multiply 38.44 by 2 to get the volume of 76.88 ft^3

You might be interested in

Could anyone help me with this question? Thank you!

shutvik [7]

Answer:

Step-by-step explanation:

We want to simplify the rational expression:

$\displaystyle \frac{n^4-11n^2+30}{n^4-7n^2+10}$

First, we can factor both the numerator and the denominator. This yields:

$\displaystyle =\frac{(n^2-6)(n^2-5)}{(n^2-2)(n^2-5)}$

Cancel like terms:

$\displaystyle =\frac{(n^2-6)}{(n^2-2)}$

However, since we canceled a term from our expression, we must place restrictions in order to stay true to the original expression.

Rational expressions are undefined whenever the denominator equals 0. The denominator is (before canceling):

$(n^2-2)(n^2-5)=0$

Solve for n. By the zero product property:

$n^2-2=0\text{ and } n^2-5=0$

Therefore:

$n^2=2\text{ and } n^2=5$

By taking the square root of both sides (since we are taking an even root, it is required to have plus/minus):

$n=\pm\sqrt{2}\text{ and } n=\pm\sqrt5$

Therefore, our fully simplified expression is:

$\displaystyle \frac{n^2-6}{n^2-2}\, ,n\neq \pm\sqrt5, \pm\sqrt2$

Hence, our answer is C!

4 0

3 years ago

Suppose that you need to create a list of n values that have a specific known mean. Some of the n values can be freely selected.

Leno4ka [110]

As the problem indicates, degrees of freedom are the number of values that can be independently selected before it is necessary to choose specific values to arrive at the desired result.

The average, on the other hand, results from the sum of a list of values divided by the amount of values in the summed list.

Assume that the mean sought is $x$ and consider that the list is composed of a single element $a$ , in that case no random number can be selected, since the mean $x$ must correspond to that number.

If the list were composed of two elements $a$ and $b$ , one of the two values could be chosen randomly, and according to the chosen value the second should be the one whose sum with the previous one results in $2x$ , this given that the formula of the average $\sum\limits_{i=1}^n \frac{ x_{i}}{n}$ .

With three values $a$ , $b$ and $c$ , it is possible to select two freely, since the thirteen must be the one that balances the sum of $a+b$ , that is $(a + b) + c = 3x$ .

Thus, in general, with n values, it is possible to select $n-1$ values freely whose sum must be balanced by the last value so that the whole sum is $nx$ .