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
For this case we have an equation of the form:
y = A * (b) ^ t
Where,
A: initial amount
b: growth rate
t: time
The given equation is:
a = 1300 (1.02) ^ 7
Where,
b = 1.02
It represents a growth of 2% on the initial amount.
Answer:
1.02 represented in this equation a growth of 2% on the initial amount.
Answer:
$18.48
Step-by-step explanation:
To find a percentage you just multiply the amount x the percentage in decimal form.
17 x 0.087 = 1.479 tax
Then you add that number too the original total.
17+1.479 = 18.479
Then you round to the nearest cent.
18.48
Total:
$18.48
Answer:
Step-by-step explanation:
LOGb(x)=y
4^5=1024
= log(4)1024=5
16 servings. Hope this helps!