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:
C.
Step-by-step explanation:
A line is straight, and a segment is a part of a line.
A and D and lines because they have the arrows on the side, meaning they keep going and are not a segment.
B is not a line, it curves.
So that leaves C to be the answer because it is the only straight one with no arrows.
Answer:
The Inequality representing money she can still spend on her friend birthday gift is 
.
Jordan can still spend at most $30 on her friends birthday gift.
Step-by-step explanation:
Given:
Total money need to spend at most = $45
Money spent on Yoga ball = $15
We need to find how much money she can still spend on her friend birthday gift.
Solution:
Let the money she can still spend on her friend birthday gift be 'x'.
So we can say that;
Money spent on Yoga ball plus money she can still spend on her friend birthday gift should be less than or equal to Total money need to spend.
framing in equation form we get;
The Inequality representing money she can still spend on her friend birthday gift is .
On solving the the above Inequality we get;
we will subtract both side by 15 using subtraction property of Inequality.
Hence Jordan can still spend at most $30 on her friends birthday gift.
20% of what is 86
0.20x = 86
x = 86/0.20
x = 430 <== weekly outcome
<em>t₁ = (3√5 + 4)/7</em>
<em>t₂ = -(3√5 - 4)/7</em>
- Step-by-step explanation:
<em>- 4 - (7t - 4)² = - 49</em>
<em>- (7t - 4)² = - 49 + 4</em>
<em>- (7t - 4) = - 45 | (-)</em>
<em>(7t - 4)² = 45</em>
<em>(7t - 4)² = 3√5</em>
<em>7t - 4 = ± 3√5</em>
- <em> 7t - 4 = 3√5 => 7t = 3√5 + 4 => t₁ = (3√5 + 4)/7</em>
- <em> 7t - 4 = - 3√5 => 7t = - 3√5 + 4 => t₂ = -(3√5 - 4)/7</em>
