M = Tim's dollar amount
R = Rebecca's dollar amount
MR = Tim and Rebecca's dollars together
R = (M x 1/10) - 6
MR = M + ((M x 1/10) - 6)
MR = 70 + ((70 x 1/10) - 6)
MR = 70 + (7 - 6)
MR = 70 + 1
MR = 71
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.
R = rate for the lawn sprinkler
rate of lawn sprinklerand hose would be 5 minutes /r ( rate per minute)
we would then want to add that to the ratio of the lawn sprinkler and hose together together which would be 5 minutes for both / 8 minutes for hose
we want to add those together to equal 100 percent, which can also be written as 1
so the correct equation would be B) 5/8 + 5/r = 1
378% is the correct answer. Divide 280 by 100 to find 1 percent. Then multiply by 135 to find 135%
Answer:
Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and observing the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.
A. The standard deviation is 0.8795. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
B. The standard deviation is 0.8795. The 10% condition is not met because there are less than 150 bass in the lake.
C. The standard deviation is 0.1204. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.
E. We are unable to determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 150 bass in the lake
The answer is E.
