The answer is 16. 7 is 1/4th the value of 28, so 4 should be 1/4th the value of 16.
Answer:
Probability that a car need to be repaired once = 20% = 0.20
Probability that a car need to be repaired twice = 8% = 0.08
Probability that a car need to be repaired three or more = 2% = 0.02
a) If you own two cars what is the probability that neither will need repair?
Probability that a car need to be repaired once , twice and thrice or more= 0.20+0.08+0.02=0.3
Probability that car need no repair = 1-0.3=0.7
Neither car will need repair=
b) both will need repair?
Probability both will need repair = 
c)at least one car will need repair
Neither car will need repair=
Probability that at least one car will need repair= 1-0.49 = 0.51
Answer:
![\large\boxed{4\sqrt[3]{64}=16}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B4%5Csqrt%5B3%5D%7B64%7D%3D16%7D)
Step-by-step explanation:
![\sqrt[3]{a}=b\iff b^3=a\\\\4\sqrt[3]{64}=(4)(4)=16\\\\\sqrt[3]{64}=4\ \text{because}\ 4^3=64](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Ba%7D%3Db%5Ciff%20b%5E3%3Da%5C%5C%5C%5C4%5Csqrt%5B3%5D%7B64%7D%3D%284%29%284%29%3D16%5C%5C%5C%5C%5Csqrt%5B3%5D%7B64%7D%3D4%5C%20%5Ctext%7Bbecause%7D%5C%204%5E3%3D64)
Answer:
Step-by-step explanation:
12P3 = 12! / (12-3)!
= 12*11*10
= 1320.
8C7 = 8!/7! *(8-7)!
= 8.
The correct answer is option c