<h2><u>
Answer: A. 15x = 10</u></h2>
Step-by-step explanation: 15 x 2/3 = 10.
-zomba
Answer:
There are 5,586,853,480 different ways to select the jury.
Step-by-step explanation:
The order is not important.
For example, if we had sets of 2 elements
Tremaine and Tre'davious would be the same set as Tre'davious and Tremaine. So we use the combinations formula.
Combinations formula:
is the number of different combinations of x objects from a set of n elements, given by the following formula.

In how many different ways can a jury of 12 people be randomly selected from a group of 40 people?
Here we have
.
So

There are 5,586,853,480 different ways to select the jury.
My first impulse is to say you don't provide enough info with which to solve this problem. We need to know how many inches in the model correspond to 1 foot in the actual plane (that is, the scale factor). What is the scale factor here?
Answer:

Step-by-step explanation:
The equation of a hyperbola centered at the origin with vertices on the y-axis is given by: 
The vertices of the hyperbola are the y-intercepts (0,12) and (0,-12)
This implies that:



The asymptote equation of a hyperbola is given by:

The given hyperbola has asymptote: 
By comparison; 


The required equation is:

Or
