Categorical(qualitative) or quantitative. Indicate whether quantitative data are continuous or discrete.

Mathematics

1 answer:

Levart [38]3 years ago

3 0

Answer:

1. Gender - qualitative

2. Age - Quantitative (Discrete)

3. Grade point Average - quantitative (Continuous )

4. Number of courses currently enrolled - quantitative (discreet)

5. Enrollment Status - qualitative

Step-by-step explanation:

Qualitative variables are those which are non-numeric, usually made up of strings or words. The quantitative variables are numeric, usually composed of digits and numbers.

Discreet variables have a finite or limited number of values which are countable as opposed to continuous variables with a non-finite range and hence, uncountable.

Gender is a qualitative variable as it is composed of non-nemeric labels which could be either Male or Female.

Age - is numeric and hence quantitative, since age is usually expressed in years, hence it is discreet.

Grade point Average is a quantitative and continuous variable due to its numeric nature and the fact that it can take up an infinite number of possible values including decimal values.

Number of courses enrolled is numeric can take a finite number of values, hence, it is discreet.

Enrollment status is qualitative as it is split into :

Part time or full time.

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Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.