Answer:
18.446
Step-by-step explanation:
Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.
To start with, we know there’s a definite amount of parking spaces in all(2,250) and we also know each level(6 in all) had 15 rows of parking spaces.
Equation
A = Parking spots per row
A = 2,250 Divided by(6 levels times 15 rows)
A= 2,250 Divided by 90
A = 25
The correct answer is 25 parking spots per row
To check this answer we can do
25 spots per row, Times 15 rows per level, Times the total amount of levels (6)
= the total amount of parking spots
25*15*6= 2,250
True
This confirms this answer as correct. Hope this helps.
Answer:
Step-by-step explanation:
