Answer:
3.4121 * 10 ^4
Step-by-step explanation:
34,121 is same as
3.4121 * 10 ^4
because the first number has to be between 0 and 10 and if you move the decimal 4 places to the right you got the initial number
Answer:
y=(x-4)²
Step-by-step explanation:
(a-b)²=a²-2ab+b²
(x-4)²=x²-2×x×4+4²
(x-4)²=x²-8x+16
Answer:
about 58 7th graders attended the dance
Step-by-step explanation:
quick maths
A is 4 and b is 63 have a good day
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.
