The eagle flies 58 miles per day. You must divide the distance (290) by the time (5 days). Since the eagles daily distance is consistant, you needn't do much more.
I believe the normal distribution is symmetrical above and below
the mean. That tells us that in this kind of distribution, the mean
and the median are the same number ... 50% of the population
is below it, and 50% is above it.
So if the female Labs' weight actually follows a 'normal' distribution
with a mean of 62.5 lbs, then 2,700 of them weigh more than 62.5 lbs
and 2,700 of them weigh less than that.
The standard deviation doesn't matter.
Answer:
f(x) = 16*(3/2)^(x-1)
Step-by-step explanation:
See attachment.
Convert the mixed numbers to the improper fractions:
![2\dfrac{3}{4}=\dfrac{2\cdot4+3}{4}=\dfrac{11}{4}\\\\19\dfrac{1}{4}=\dfrac{19\cdot4+1}{4}=\dfrac{77}{4}](https://tex.z-dn.net/?f=2%5Cdfrac%7B3%7D%7B4%7D%3D%5Cdfrac%7B2%5Ccdot4%2B3%7D%7B4%7D%3D%5Cdfrac%7B11%7D%7B4%7D%5C%5C%5C%5C19%5Cdfrac%7B1%7D%7B4%7D%3D%5Cdfrac%7B19%5Ccdot4%2B1%7D%7B4%7D%3D%5Cdfrac%7B77%7D%7B4%7D)
![2\dfrac{3}{4}a=19\dfrac{1}{4}\\\\\dfrac{11}{4}a=\dfrac{77}{4}\qquad\text{multiply both sides by 4}\\\\\not4^1\cdot\dfrac{11}{\not4_1}a=\not4^1\cdot\dfrac{77}{\not4_1}\\\\11a=77\qquad\text{divide both sides by 11}\\\\\boxed{a=7}](https://tex.z-dn.net/?f=2%5Cdfrac%7B3%7D%7B4%7Da%3D19%5Cdfrac%7B1%7D%7B4%7D%5C%5C%5C%5C%5Cdfrac%7B11%7D%7B4%7Da%3D%5Cdfrac%7B77%7D%7B4%7D%5Cqquad%5Ctext%7Bmultiply%20both%20sides%20by%204%7D%5C%5C%5C%5C%5Cnot4%5E1%5Ccdot%5Cdfrac%7B11%7D%7B%5Cnot4_1%7Da%3D%5Cnot4%5E1%5Ccdot%5Cdfrac%7B77%7D%7B%5Cnot4_1%7D%5C%5C%5C%5C11a%3D77%5Cqquad%5Ctext%7Bdivide%20both%20sides%20by%2011%7D%5C%5C%5C%5C%5Cboxed%7Ba%3D7%7D)