The question might have some mistake since there are 2 multiplier of t. I found a similar question as follows:
The population P(t) of a culture of bacteria is given by P(t) = –1710t^2+ 92,000t + 10,000, where t is the time in hours since the culture was started. Determine the time at which the population is at a maximum. Round to the nearest hour.
Answer:
27 hours
Step-by-step explanation:
Equation of population P(t) = –1710t^2+ 92,000t + 10,000
Find the derivative of the function to find the critical value
dP/dt = -2(1710)t + 92000
= -3420t + 92000
Find the critical value by equating dP/dt = 0
-3420t + 92000 = 0
92000 = 3420t
t = 92000/3420 = 26.90
Check if it really have max value through 2nd derivative
d(dP)/dt^2 = -3420
2nd derivative is negative, hence it has maximum value
So, the time when it is maximum is 26.9 or 27 hours
To solve for the confidence interval for the population
mean mu, we can use the formula:
Confidence interval = x ± z * s / sqrt (n)
where x is the sample mean, s is the standard deviation,
and n is the sample size
At 95% confidence level, the value of z is equivalent to:
z = 1.96
Therefore substituting the given values into the
equation:
Confidence interval = 3 ± 1.96 * 5.8 / sqrt (51)
Confidence interval = 3 ± 1.59
Confidence interval = 1.41, 4.59
Therefore the population mean mu has an approximate range
or confidence interval from 1.41 kg to 4.59 kg.
(x) = x – 2inverse of the function
Yes. I think so. Even if there is not intercept the intercept is 0
