Answer:
By dividing the top number by the bottom.
Step-by-step explanation:
ummmmm
Answer:
75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.
Step-by-step explanation:
The question is missing. It is as follows:
Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69 104 125 129 60 64
Assume that the population of x values has an approximately normal distribution.
Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)
75% Confidence Interval can be calculated using M±ME where
- M is the sample mean weight of the wild mountain lions (
)
- ME is the margin of error of the mean
And margin of error (ME) of the mean can be calculated using the formula
ME=
where
- t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)
- s is the standard deviation of the sample(31.4)
Thus, ME=
≈16.66
Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5
Answer:
See explanation.
(Before continuing reading, I took the base to be 3. Please tell me if you didn't want the base to be 3.)
Step-by-step explanation:
I assume 3 is suppose to be the base. Let's list some values that can be written as 3 to some integer.
3^0=1
3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
......
I could have also did negative integer powers, but this is all I really need to convince you that log_3(28) is between 3 and 4.
log_3(28) means the value x such that 3^x=28.
Since 28 is between 27 and 81 in my list above, that means 3^x is between 3^3 and 3^4. This means that x is a value between 3 and 4.
Answer:
2ax − 6ay + bx − 3
Step-by-step explanation:
The expression is not factorable with rational numbers
Answer:
-51.4
Step-by-step explanation: