Correct answer is: (0,7843) and (10,8793)
Solution:-
Given that a junior college has an enrollment of 7843 students in 1990 and 8793 students in year 2000.
We have to write this data as (x,y) .
Where x= years after 1990 and y=number of students enrolled.
Since in 1990, 7843 students enrolled, x = 1990-1990=0
And y=7843.
Hence one ordered pair is (0,7843).
Let us find the years after 1990 for 2000 = 2000-1990 =10
Hence another ordered pair is (10,8793).
The answer is sixteen thousand four hundred ninety
Three times. The remainder would be 2.
Answer:
The mean number of adults who would have bank savings accounts is 32.
Step-by-step explanation:
For each adult surveyed, there are only two possible outcomes. Either they have bank savings accounts, or they do not. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:

In this problem, we have that:

If we were to survey 50 randomly selected adults, find the mean number of adults who would have bank savings accounts.
This is E(X) when
.
So

The mean number of adults who would have bank savings accounts is 32.
Answer:
228.42 ft
Step-by-step explanation:
To solve for this question, we would be applying the Trigonometric function of Sine.
Sin theta = Opposite/Hypotenuse
Theta = 52°
Opposite = 180ft
Hypotenuse = Length of the wire = x
Hence:
sin 52 = 180/x
x = 180/sin 52
x = 228.42327871 ft
Approximately = 228.42 ft