Jason --> 126/6 = 21 pages per hour
joe --> 115/5 = 23 pages per hour
joe reads 2 more pages per hour
hope this helps :)
Answer:
C
Step-by-step explanation:
Multiply the 1.5 hours per pound by 2.75 pounds and we get 4.125 hours or 4 1/8
Answer:
a) p-hat (sampling distribution of sample proportions)
b) Symmetric
c) σ=0.058
d) Standard error
e) If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).
Step-by-step explanation:
a) This distribution is called the <em>sampling distribution of sample proportions</em> <em>(p-hat)</em>.
b) The shape of this distribution is expected to somewhat normal, symmetrical and centered around 16%.
This happens because the expected sample proportion is 0.16. Some samples will have a proportion over 0.16 and others below, but the most of them will be around the population mean. In other words, the sample proportions is a non-biased estimator of the population proportion.
c) The variability of this distribution, represented by the standard error, is:
d) The formal name is Standard error.
e) If we divided the variability of the distribution with sample size n=90 to the variability of the distribution with sample size n=40, we have:

If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).
Answer:115 apex
Step-by-step explanation:
Answer:
78% probability that a randomly selected online customer does not live within 50 miles of a physical store.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this problem, we have that:
Total outcomes:
100 customers
Desired outcomes:
A clothing vendor estimates that 78 out of every 100 of its online customers do not live within 50 miles of one of its physical stores. So the number of desired outcomes is 78 customers.
Using this estimate, what is the probability that a randomly selected online customer does not live within 50 miles of a physical store?

78% probability that a randomly selected online customer does not live within 50 miles of a physical store.