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4 years ago

(03.01) Is the following relation a function? y equals the sine of 3 x

Mathematics

2 answers:

Illusion [34]4 years ago

7 0

Answer:

Yes. This is a function.

Step-by-step explanation:

Given is a function as

$y = sin 3x$

We find that for every x in the domain of real numbers there is a unique image for every x as sin 3x.

Thus this relation given satisfies the requirements of a function.

Also we find that this function is periodic with amplitude 1 and period = 2pi/3

So vertical line test if done will find that there is only one intersection for every vertical line throughout the domain

Hence this relation is definitely a function

nikdorinn [45]4 years ago

6 0

A relation is a function if it passes the vertical line test. That graph in a coordinate plane is just a sin wave that is spread out. If you run a perfectly vertical line through the graph at every place on the graph, the line you draw will only go through the graph at one place every time. That makes it a function. It's not a one-to-one function cuz it fails the horizontal line test, but it is a function!

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3 years ago

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Semmy [17]

Answer:

Step-by-step explanation:

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3 years ago

Suppose that 7 in every 10 auto accidents involve a single vehicle. If 15 auto accidents are randomly selected, compute the prob

kobusy [5.1K]

Answer:

$\mathbf{P(X \le 4 ) \simeq 0.0006722}$

Step-by-step explanation:

From the information given:

p = x/n

p = 7/10

p = 0.7

sample size n = 15

Suppose X be the number of accidents involved by a single-vehicle.

Then;

$X \sim Binom (15,0.7)$

Thus, the required probability that at most 4 involve in a single-vehicle is

$P(X\le 4) \\ \\P(X \le 4) = P(X = 0) + P(X =1 ) + ... + P(X = 4)$

$P(X \le 4 ) = (^{15}_0) *0.7^0 *0.3^{15-0} + (^{15}_1) *0.7^1 *0.3^{15-1} + (^{15}_2) *0.7^2 *0.3^{15-2} + (^{15}_3) *0.7^3 *0.3^{15-3} + (^{15}_4) *0.7^4 *0.3^{15-4}$

$P(X \le 4 ) = (\dfrac{15!}{0!(15-0)!}) *0.7^0 *0.3^{15-0} + (\dfrac{15!}{1!(15-1)!}) *0.7^1 *0.3^{15-1} + (\dfrac{15!}{2!(15-2)!}) *0.7^2 *0.3^{15-2} + (\dfrac{15!}{3!(15-3)!}) *0.7^3 *0.3^{15-3} + (\dfrac{15!}{4!(15-4)!}) *0.7^4 *0.3^{15-4}$

$P(X \le 4 ) =1.4348907 \times 10^{-8} +5.02211745 \times 10^{-7} + 8.20279183 \times 10^{-6} + 8.29393397 \times 10^{-5} + 5.80575378 \times 10^{-4}$

$P(X \le 4 ) =6.7223407 \times 10^{-4}$

$\mathbf{P(X \le 4 ) \simeq 0.0006722}$

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3 years ago