For this case we have an equation of the form:
y = A * (b) ^ t
Where,
A: initial amount
b: decrease factor
t: time
Substituting values:
y = 30 * (b) ^ t
To calculate t we use a point in the table.
We have:
(t, y) = (1, 28.5)
Substituting:
28.5 = 30 * (b) ^ 1
b = 28.5 / 30
b = 0.95
Answer:
a. Decay factor is 0.95
Answer:
40x^6
Step-by-step explanation: took the test
77 times 100 is 7,700. So your answer is 7,700
Answer:
Step-by-step explanation:
We want to determine a 90% confidence interval for the mean amount of time that teens spend online each week.
Number of sample, n = 41
Mean, u = 43.1 hours
Standard deviation, s = 5.91 hours
For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean +/- z ×standard deviation/√n
It becomes
43.1 ± 1.645 × 5.91/√41
= 43.1 ± 1.645 × 0.923
= 43.1 ± 1.52
The lower end of the confidence interval is 43.1 - 1.52 =41.58
The upper end of the confidence interval is 43.1 + 1.52 =44.62
Therefore, with 90% confidence interval, the mean amount of time that teens spend online each week is between 41.58 and 44.62
Answer:
2nd quadrant.
Step-by-step explanation:
