Answer:
.6
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.
Answer:
The second one if that how to answer.
Step-by-step explanation:
S = πr(r + √(h² + r²))
400.2 = 3.14(6)(6 + √(h² + 6²))
400.2 = 18.84(6 + √(h² + 36))
18.84 18.84
21¹⁰⁹/₄₇₁ = 6 + √(h² + 36))
- 6 - 6
15¹⁰⁹/₄₇₁ = √(h² + 36)
231²²¹⁰⁰⁵/₂₂₁₈₄₁ = h² + 36
- 36 - 36
195²²¹⁰⁰⁵/₂₂₁₈₄₁ = h²
14 ≈ h
