Answer: A Rhombus has 4 congruent sides .
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.
Answer:
x∈R
Step-by-step explanation:
Function rule is the relationship between input and output values. To find the function rule we have to observe the given data carefully that how input and output values are related to each other.
I believe the correct answer is true. <span>When solving a system of linear equations, try to algebraically form one equation that has only one variable. In this way, you can solve the value of that variable and eventually solve the other variables. Hope this answers the question. Have a nice day.</span>
