P(t) = [ 64/ (1 + 11.e⁽⁻⁰ ⁰⁸t⁾]
In 1991, t = 1, hence:
P(t) = [ 64/ (1 + 11.e⁽⁻⁰ ⁰⁸ˣ¹⁾] = 5.7377 billion or rounded 5.74 billion
(Answer A)
Answer:
σ should be adjusted at 0.5.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
Approximately 68% of the measures are within 1 standard deviation of the mean.
Approximately 95% of the measures are within 2 standard deviations of the mean.
Approximately 99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean 12.
Assuming we can precisely adjust σ, what should we set σtobe so that the actual amount dispensed is between 11 and 13 ounces, 95% of the time?
13 should be 2 standard deviations above the mean of 12, and 11 should be two standard deviations below the mean.
So 1 should be worth two standard deviations. Then



σ should be adjusted at 0.5.
can you please rotate the picture
Answer:
48 students
Step-by-step explanation:
multpily 75 by 64%. 0.64*75= 48/
I think it would be Garo, Annette, and Heather????