Answer:
84
Step-by-step explanation:
because 72 girl than 84 girls is greater
We havep(X)=eβ0+β1X1+eβ0+β1X⇔eβ0+β1X(1−p(X))=p(X),p(X)=eβ0+β1X1+eβ0+β1X⇔eβ0+β1X(1−p(X))=p(X),which is equivalent top(X)1−p(X)=eβ0+β1X.p(X)1−p(X)=eβ0+β1X.
To use the Bayes classifier, we have to find the class (kk) for whichpk(x)=πk(1/2π−−√σ)e−(1/2σ2)(x−μk)2∑Kl=1πl(1/2π−−√σ)e−(1/2σ2)(x−μl)2=πke−(1/2σ2)(x−μk)2∑Kl=1πle−(1/2σ2)(x−μl)2pk(x)=πk(1/2πσ)e−(1/2σ2)(x−μk)2∑l=1Kπl(1/2πσ)e−(1/2σ2)(x−μl)2=πke−(1/2σ2)(x−μk)2∑l=1Kπle−(1/2σ2)(x−μl)2is largest. As the log function is monotonally increasing, it is equivalent to finding kk for whichlogpk(x)=logπk−(1/2σ2)(x−μk)2−log∑l=1Kπle−(1/2σ2)(x−μl)2logpk(x)=logπk−(1/2σ2)(x−μk)2−log∑l=1Kπle−(1/2σ2)(x−μl)2is largest. As the last term is independant of kk, we may restrict ourselves in finding kk for whichlogπk−(1/2σ2)(x−μk)2=logπk−12σ2x2+μkσ2x−μ2k2σ2logπk−(1/2σ2)(x−μk)2=logπk−12σ2x2+μkσ2x−μk22σ2is largest. The term in x2x2 is independant of kk, so it remains to find kk for whichδk(x)=μkσ2x−μ2k2σ2+logπkδk(x)=μkσ2x−μk22σ2+logπkis largest.
ng expression
∫0.950.0510dx+∫0.050(100x+5)dx+∫10.95(105−100x)dx=9+0.375+0.375=9.75.∫0.050.9510dx+∫00.05(100x+5)dx+∫0.951(105−100x)dx=9+0.375+0.375=9.75.So we may conclude that, on average, the fraction of available observations we will use to make the prediction is 9.75%9.75%.res. So when p→∞p→∞, we havelimp→∞(9.75%)p=0.
Some fractions that are equal to 2/3 are: 4/6 and 6/9
(You just have to multiply the numerator and denominator by the same number.)
Decimal: 0.666...
(Divide the numerator by the denominator.)
Percent: 66.6%
(Multiply the decimal by 100.)
<em>Hope it helps</em> :)
Answer:
amenda. de ce ai nevoie?
Step-by-step explanation:
Answer:
A. 11
Step-by-step explanation:
Since the angles are vertical angles (share a vertex) they are equal in measure.
160 = 10x + 50
subtract 50 from both sides
110 = 10x
divide both sides by 10
x = 11
