Answer:
81.85%
Step-by-step explanation:
Given :
The average summer temperature in Anchorage is 69°F.
The daily temperature is normally distributed with a standard deviation of 7°F .
To Find:What percentage of the time would the temperature be between 55°F and 76°F?
Solution:
Mean = 
Standard deviation = 
Formula : 
Now At x = 55


At x = 76


Now to find P(55<z<76)
P(2<z<-1)=P(z<2)-P(z>-1)
Using z table :
P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185
Now percentage of the time would the temperature be between 55°F and 76°F = 
Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.
Answer:
A: P = 41100 - 590t
B: 34610
Step-by-step explanation:
equation = 41100-(41100-36380)/(2011-2003)t
plug it in
41100 - 590(2014-2003)=34610
Answer:
cool
Step-by-step explanation:
Answer:
Range = (237100, 292900)
Step-by-step explanation:
Using Chebyshevs Inequality:






Thus, 88.9% of the population is within 3 standard deviation of the mean with the Range = μ ± kσ
where;
μ = 265000
σ = 9300
Range = 265000 ± 3(9300)
Range = 265000 ± 27900
Range = (265000 - 27900, 265000 + 27900)
Range = (237100, 292900)