Answer:
5/9
Step-by-step explanation:
Answer:
C. Test for Goodness-of-fit.
Step-by-step explanation:
C. Test for Goodness-of-fit would be most appropriate for the given situation.
A. Test Of Homogeneity.
The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.
B. Test for Independence.
The chi square is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.
C. Test for Goodness-of-fit.
The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.
I've answered your other question as well.
Step-by-step explanation:
Since the identity is true whether the angle x is measured in degrees, radians, gradians (indeed, anything else you care to concoct), I’ll omit the ‘degrees’ sign.
Using the binomial theorem, (a+b)3=a3+3a2b+3ab2+b3
⇒a3+b3=(a+b)3−3a2b−3ab2=(a+b)3−3(a+b)ab
Substituting a=sin2(x) and b=cos2(x), we have:
sin6(x)+cos6(x)=(sin2(x)+cos2(x))3−3(sin2(x)+cos2(x))sin2(x)cos2(x)
Using the trigonometric identity cos2(x)+sin2(x)=1, your expression simplifies to:
sin6(x)+cos6(x)=1−3sin2(x)cos2(x)
From the double angle formula for the sine function, sin(2x)=2sin(x)cos(x)⇒sin(x)cos(x)=0.5sin(2x)
Meaning the expression can be rewritten as:
sin6(x)+cos6(x)=1−0.75sin2(2x)=1−34sin2(2x)
Answer: 2^14
Step-by-step explanation: