42 is the answer
merry Christmas and a happy new year :)
Answer:
Y=⅘X+2. the picture not clear so i hesitate
Answer:
The correct option is (b).
Step-by-step explanation:
If X 
N (µ, σ²), then 
, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z N (0, 1).
The distribution of these z-variate is known as the standard normal distribution.
The mean and standard deviation of the active minutes of students is:
<em>μ</em> = 60 minutes
<em>σ </em> = 12 minutes
Compute the <em>z</em>-score for the student being active 48 minutes as follows:
Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.
The correct option is (b).
Answer:
I should use at least 304 students
Step-by-step explanation:
Margin error (E) = t × sd/√n
E = 40
sd = 300
confidence level (C) = 98% = 0.98
significance level = 1 - C = 1 - 0.98 = 0.02 = 2%
t-value corresponding to 2% significance level and infinity degree of freedom is 2.326
n = (t×sd/E)^2 = (2.326×300/40)^2 = 17.445^2 = 304 (to the nearest integer)
It would be 3.87,to the nearest hundred cause the real length is √15
