9.9^2X1.79
9.9^2=98.01
98.01X1.79=175.4379
Answer:
The answer to your question is the theater is 6.7 mi from home
Step-by-step explanation:
Data
speed 1 = 2 mi/h
time in theater = 2h 15 min
speed 2 = 40 mi/h
time = 3.5 h
Process
1.- Write the formula to calculate speed
speed = distance / time
-Solve for time
time = speed x time
2.- Write and equation to solve this problem
total time = time from home to theater + time from theater to home
3.- Substitution
3.5 = d/2 + d/40
4.- Solve for d
3.5 = (20d + d) / 40
140 = 21d
d = 140 / 21
d = 6.7 mi
Answer:
See Below
Step-by-step explanation:
The function is a piecewise function defined as:
a)
We need to find the limit of the function as t goes to infinity. This means what is the max value of fish in the pond given times goes to infinity (on an on).
We will take the 2nd part of the equation since t falls into that range, t is infinity, which is definitely greater than 8.
This means the maximum number of fish at this pond is 1600, no matter how long it goes on.
b)
A function is continuous at a point if we have the limit and the functional value at that point same.
Functional value at t = 8 is (we use 2nd part of equation):
We do have a value and limit also goes to this as t approaches 8.
So, function is continuous at t = 8
c)
We want to find is there a "time" when the number of fishes in the pond is 250, during t from 0 to 6. We plug in 250 into N(t) and try to find t. Make sure to use the 1st part of the piece-wise function. Shown below:
The time is 4 years when the number of fishes in the pond is 250
Answer:
99.85%
Step-by-step explanation:
The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives 10.4 years; the standard deviation is 1.9 years.
Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living less than 16.1 years.
Solution:
The empirical rule states that for a normal distribution most of the data fall within three standard deviations (σ) of the mean (µ). That is 68% of the data falls within the first standard deviation (µ ± σ), 95% falls within the first two standard deviations (µ ± 2σ), and 99.7% falls within the first three standard deviations (µ ± 3σ).
Therefore:
68% falls within (10.4 ± 1.9). 68% falls within 8.5 years to 12.3 years
95% falls within (10.4 ± 2*1.9). 95% falls within 6.6 years to 14.2 years
99.7% falls within (10.4 ± 3*1.9). 68% falls within 4.7 years to 16.1 years
Probability of a meerkat living less than 16.1 years = 100% - (100% - 99.7%)/2 = 100% - 0.15% = 99.85%
