Does the graph represent a function? Khan Academy

Mathematics

2 answers:

Ierofanga [76]2 years ago

7 0

Answer:

yes

Step-by-step explanation:

try using a vertical line, if you move the vertical line across the graph and the line touches the graph only at one point, than it means it is a function

serg [7]2 years ago

6 0

Answer:

This graph represents a function

Step-by-step explanation:

By its definition, a function is a relation or a correlation between two sets (groups) of values where every input corresponds to only one output. What this means is that every input in a function can only have one output. This characteristic results in a unique property in the graph of a function, that is, the verticle line test. The verticle line test states that when one draws a line through the graph of a function, the line can only intersect the graph in exactly one place in the function. If the line intersects the graph in more than one place, then the graph does not represent a function. In the given situation, any verticle line drawn on the graph only intersects the graph in one place, thus the graph represents a function.

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evablogger [386]

A.) For n independent variates with the same distribution, the standard deviation of their mean is the standard deviation of an individual divided by the square root of the sample size: i.e. s.d. (mean) = s.d. / sqrt(n)
Therefore, the standard deviation of of the average fill volume of 100 cans is given by 0.5 / sqrt(100) = 0.5 / 10 = 0.05

b.) In a normal distribution, P(X < x) is given by P(z < (x - mean) / s.d).
Thus, P(X < 12) = P(z < (12 - 12.1) / 0.05) = P(z < -2) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275

c.) Let the required mean fill volume be u, then P(X < 12) = P(z < (12 - u) / 0.05) = 1 - P(z < (u - 12) / 0.05) = 0.005
P(z < (u - 12) / 0.05) = 1 - 0.005 = 0.995 = P(z < 2.575)
(u - 12) / 0.05 = 2.575
u - 12 = 2.575 x 0.05 = 0.12875
u = 12 + 0.12875 = 12.12875
Therefore, the mean fill volume should be 12.12875 so that the probability that the average of 100 cans is below 12 fluid ounces be 0.005.

d.) Let the required standard deviation of fill volume be s, then P(X < 12) = P(z < (12 - 12.1) / s) = 1 - P(z < 0.1 / s) = 0.005
P(z < 0.1 / s) = 1 - 0.005 = 0.995 = P(z < 2.575)
0.1 / s = 2.575
s = 0.1 / 2.575 = 0.0388
Therefore, the standard deviation of fill volume should be 0.0388 so that the probability that the average of 100 cans is below 12 fluid ounces be 0.005.

e.) Let the required number of cans be n, then P(X < 12) = P(z < (12 - 12.1) / (0.5/sqrt(n))) = 1 - P(z < (12.1 - 12) / (0.5/sqrt(n))) = 0.01
P(z < 0.1 / (0.5/sqrt(n))) = 1 - 0.01 = 0.99 = P(z < 2.327)
0.1 / (0.5/sqrt(n)) = 2.327
0.5/sqrt(n) = 0.1 / 2.327 = 0.0430
sqrt(n) = 0.5/0.0430 = 11.635
n = 11.635^2 = 135.37
Therefore, the number of cans that need to be measured such that the average fill volume is less than 12 fluid ounces be 0.01

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