Hmmmt that’s a good question
Answer:
$2.79 per pound of apples.
0.36 pounds per dollar.
The unit rate representing cost of per pound apples is typically used.
Step-by-step explanation:
We have been given that the weekly ad for a local grocery store advertises a 5 pound bag of organic apples for $13.95.
Since we know when rates are expressed as a quantity of 1, such as 100 meter per second or 5 miles per hour, they are called unit rates. We need to have 1 in our denominator to express two quantities as unit rate.
Let us find the unit rate in terms of cost of per pound bag of apples.


Therefore, our unit rate will be $2.79 per pound of apples.
Let us find unit rate of pounds of apples per dollar.



Therefore, our another unit rate will be 0.36 pounds per dollar.
Since, we purchase apples or other items according to their price per piece or their price per pound in our daily life, therefore, the unit rate representing cost of per pound apples is typically used.
Answer: f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.
Step-by-step explanation:
Given function: 
The vertical asymptote occurs for those values of x which make function indeterminate or denominator 0.
i.e. 
Hence, f(x) will have vertical asymptotes at x=-2 and x=2.
To find the horizontal asymptote , we can see that the degree of numerator and denominator is same i.e. 2.
So, the graph will horizontal asymptote at 
i.e. 
Hence, f(x) will have horizontal asymptote at y=3.
Where are the sets? I don't see any...
Answer:
The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.
Step-by-step explanation:
We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.
If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.
The standard deviation of the sample is equal to:

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:
