Answer:
Not a function.
Does not pass vertical line test.
B.
Step-by-step explanation:
Something called a vertical line test is used to determine if a relation that has been graphed is a function or not.
We say if it passes then it is a function.
It will pass if you are able to draw infinitely many vertical lines covering the whole graph and each vertical line either touches your relation once or none.
If a single vertical line that you draw touches more than once, then it isn't a function.
When I say draw, I don't mean you should physically do it, but more so imagine it.
Now this particular relation is not a function because I can find a vertical line that touches more than once. Take the vertical line x=5 for example.
It will touch at (5,-6) and (5,6). You cannot have an x assigned to more than one y.
Answer:
38 students
Step-by-step explanation:
251 students minus the students traveled in cars =228 divide that by six buses and your outcome will be 38 students
Answer: no, the function has no real number zeros.
Step-by-step explanation:
Answer:
C. -4x - 1
Step-by-step explanation:

Distribute the term outside of the parenthesis, Multiply each term in the parenthesis by the term outside (do not forget to include the negative sign when multiplying);

Simplify,

Answer:
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = 1.1
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given total number of students n(T) = 150
Given 125 of them are fluent in Swahili
Let 'S' be the event of fluent in Swahili language
n(S) = 125
The probability that the fluent in Swahili language

Let 'E' be the event of fluent in English language
n(E) = 135
The probability that the fluent in English language

n(E∩S) = 95
The probability that the fluent in English and Swahili

<u><em>Step(ii):</em></u>-
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = P(S) + P(E) - P(S∩E)
= 0.833+0.9-0.633
= 1.1
<u><em>Final answer:-</em></u>
The probability that a student chosen at random is fluent in English or Swahili.
P(S∪E) = 1.1