The reciprocal of 3 is 1/3.
Answer:

Step-by-step explanation:

by multiplying

Answer: -8.92
Step-by-step explanation:
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
Step-by-step explanation:
x = number hours dog walking
y = number hours tutoring
x + y <= 11
6x + 10y >= 80
the solution is the common area below the first line and above the second line.
the first line is above the second line (due to the y-intercept of 11 vs. 8) until the crossing point.
for the crossing point we use the regular equations :
from the first we get
x = 11 - y
using this in the second
6×(11 - y) + 10y = 80
66 - 6y + 10y = 80
4y = 14
y = 14/4 = 7/2 = 3.5
x = 11 - y = 11 - 3.5 = 7.5
so, she has to do at least 3.5 hours tutoring (and then 7.5 hours dog walking) for the minimum of $80 earning, all the way up to the maximum of 11 hours tutoring (and 0 dog walking) for $110 earning.