Answer:
2.404 x 10^6?. A. the population of Iowa. B. the diameter (in meters) of an atom. C. the thickness (in meters) of a sheet of printer paper. D. the amount of money (in US dollars) in the average checking account. .
When 2.404 x 10^6 is expressed in another form, the number is equal to 2,404,000. The population of a place would possibly be around this number. It cannot be the diameter of an atom since the number is too large not the thickness of a paper. It is not a common case also that the money in a bank account is around this number. Answer is A.
Answer:
Step-by-step explanation:
The ratio of the angle is how the angle relates to each other. Let's say that we try to relate Angle A, B, C and get 1 : 3 : 5. That means Angle B is three times Angle A and Angle C is 5 times Angle A
Thus we get the equation:
A
B = 3A
C = 5A
A + B + C = 180, since all the angles of the triangle is equal to 180 degrees
A + 3A + 5A = 180
9A = 180
A = 20 --> B = 3A = 60 --> C = 5A = 100
So Angle A is 20 degrees, Angle B is 60 degrees, and Angle C is 100 degrees.
Hope that helps!
Answer:
Step-by-step explanation:you take the 740 and divide it by 40 to get the hourly wage then times that by the 32 and theirs your answer
Answer:
The probability is 
Step-by-step explanation:
We can divide the amount of favourable cases by the total amount of cases.
The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, 
For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function 
with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.
Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.
We can conclude that the probability for 8 rooks not being able to capture themselves is
