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lorasvet [3.4K]
3 years ago
11

Vhich of the following functions have graphs that oscillate? Select all that apply

Mathematics
1 answer:
gladu [14]3 years ago
6 0

Answer:

Only options A and B are correct, i.e. y = sin∅ and y = cos∅.

Step-by-step explanation:

The graphs that oscillate, should be periodic in nature and continuous functions.

We know all trigonometric functions are periodic in nature. But not all trigonometric functions are continuous.

The function is continuous if its values are defined at all angles and it should not be undefined for any particular angle.

We know the following facts:-

tan(90°) = ∞

sec(90°) = ∞

csc(0°) = ∞

cot(0°) = ∞

<u>Only sin and cos don't have values ∞ at any angle.</u> It means only sin and cos are continuous functions.

Hence, only options A and B are correct, i.e. y = sin∅ and y = cos∅.

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Give the line that is parallel to the line y=-5x-2 and goes through (3,-9)
igor_vitrenko [27]

Answer: I don’t think there is such a line. A Line that passes through the point (3,-9) must have a positive slope, unlike the line y=-5x-2, which has a negative slope. If a line where to go through (3,-9), is would intersect the line you presented.

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2 years ago
Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

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Answer: 10x^2+3xy+6x-y^2+3y

Step-by-step explanation:

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Then, knowing this property, now you have to apply the Distributive property. So:

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Finally, to simplify this expression, you need to add the like terms.

Therefore, you get that the product is:

=10x^2+3xy+6x-y^2+3y

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