Well you're given the equation and the time.
substitute:
V(t) = 26,000(0.90^t)
V(11) = 26,000(0.90^(11))
V(11) = 8,159.075498
V(11) ≈ 8,159
Answer:
45cm
Step-by-step explanation:
Volume of a sphere = 4/3πr³
Volume of an hemisphere = 1/2 of volume of a sphere
Volume of hemisphere = 1/2{4/3πr³}
Volume of hemisphere = 2/3πr³
Given
Volume of the hemisphere = 60,750πcm³
60,750π = 2/3πr³
π cancels out
60,750 = 2/3r³
Multiply both sides by 3
2r³ = 60,750×3
2r³=182,250
Divide both sides by 2
r³ = 91,125
Find the ∛ of 91,125 to eliminate the cube on r
r = ∛91,125
r = 45cm
Its 1 quattordecillion or 1(48 zeros that I don't want to type)
Answer: 0.4512
Step-by-step explanation:
A bit string is sequence of bits (it only contains 0 and 1).
We assume that the 0 and 1 area equally likely to any place.
i.e. P(0)= P(1)= 
The length of bits : n = 10
Let X = Number of getting ones.
Then , 
Binomial distribution formula :
, where p= probability of getting success in each event and q= probability of getting failure in each event.
Here , 
Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.






Hence, the probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.