A cube 4 units on each side is composed of 64 unit cubes. Two faces of the larger cube that share an edge are painted blue, and the cube is disassembled into 64 unit cubes. Two of the unit cubes are selected uniformly at random. What is the probability that one of two selected unit cubes will have exactly two painted faces while the other unit cube has no painted faces?

P = 0.0714

Step-by-step explanation:

If two faces of the larger cube that share and edge are painted blue, it means that 28 of the 64 unit cubes are painted in at least one side and 36 cubes have no painting faces.

Additionally, from the 28 cubes painted only 4 have exactly two painted faces.

Then, to calculate the number of ways in which we can select x elements from a group of n, we can use the following equation:

So, the probability that one of two selected unit cubes will have exactly two painted faces while the other unit cube has no painted faces is:

Because there are 64C2 ways to select 2 cubes from the 64, and from that, there are 4C1*36C1 ways to select one cube with exactly two painted faces and one cube with no painted faces.

$655.61 which is option c

step-by-step explanation:

price of house =   $146,000

initial rate = 3.5%

mortgaged paid in this = 3.5 % of $146,000

                                        =

                                        =$5110

the total years would be = 7

total mortgaged paid = 7 * $5110

                                      =$35770

remaining money = $146000-35770

                                =$110230

new mortgaged payment = 5 % of 110230

                                            =  

                                            =$655.61

d. 145*2 is 290 than the total angle is 360 then minus 290 then divide by two