Distance to the star in light years = 3 * 0.62 * 9.5 * 10^12 = 1.767 * 10^13 miles
dividng this by 17,000 gives 1.039411765 * 10^9 hours
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:
2x -5, like that you didn't put enough.
Hi, P = l+l+w+w 520-120-120= 280÷2 w = 140ft
Answer:
![y=2sin[\frac{\pi}{4}(x+3) ]-5](https://tex.z-dn.net/?f=y%3D2sin%5B%5Cfrac%7B%5Cpi%7D%7B4%7D%28x%2B3%29%20%5D-5)
Step-by-step explanation:
The standard form for a sinusoidal function is given by:
y = Asin[B(x - C)] + D
Where A is the amplitude of the sine wave, the period is 2π/B, C is the phase shift and D is the vertical displacement / translation
Given:
A = 2, D = -5 (down), C = -3 (left).
8 = 2π/B; B = π/4
Therefore:
![y=2sin[\frac{\pi}{4}(x+3) ]-5](https://tex.z-dn.net/?f=y%3D2sin%5B%5Cfrac%7B%5Cpi%7D%7B4%7D%28x%2B3%29%20%5D-5)