Answer:
Max number of bags is 16.
Step-by-step explanation:
48 - cc
64- v
100 - rc
= 21
= 33,(3)
Max number of bags is 16.
What is the product? (-3s+2t)(4s-t)
2 answers:
You might be interested in
(a) X ≈ N (0.10 , 0.02²)
(b) The probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.
(c) The 60th percentile is 11%.
<h3>The complete question is:</h3><h3>The average THC content of marijuana sold on the street is 10%. Suppose the THC content is normally distributed with standard deviation of 2%. Let X be the THC content for a randomly selected bag of marijuana that is sold on the street. Round all answers to two decimal places and give THC content in units of percent. For example, for a THC content of 11%, write "11" not 0.11. </h3>A. X ~ N( , )
B. Find the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11%.
C. Find the 60th percentile for this distribution.
According to the question,
The random variable X is defined as the THC content for a bag of marijuana that is sold on the street.
(a)The random variable X is Normally distributed with mean, μ = 0.10 and standard deviation, σ = 0.02.
Therefore, X ≈ N (0.10 , 0.02²)
(b) The value of P (X > 0.11) as follows:
P (X > o.11_ = P(X -μ/a > )
= P(Z > 0.50)
= 1 -P (Z <0.50)
= 1- 0.69146
= 0.30854 ≈ 0.31
Hence,
the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.
(c)Let x represent the 60th percentile value.
Then, P (X < x) = 0.60.
⇒ P (Z < z) = 0.60
The value of z is,
z = 0.26.
Now put the value of x as follows:
z = x - μσ
0.26 = x - 0.100.02
x = 0.10+ ( 0.26 x 0.02)
x = 0.1052
x ≈ 0.11
Therefore,
The 60th percentile is 11%
Learn more about Probability here:
#SPJ1