Answer:
Can you please give more information about the question?
Step-by-step explanation:
:
Answer:
i think 1, 3, 4 for rational
the rest for irrational
Step-by-step explanation:
i'm sorry if it's wrong >_<
Answer : $36
Explanation:
To find the sale price we can set up a proportion. 25/100 = x/48. We keep the 100 and 48 both on the bottom of the fraction since they represent the "whole". x represents the discount off the full price of the dress. To solve this proportion, we cross multiply, yielding 48 * 25 = 100x. Alternatively, if you notice that 25/100 simplifies to 1/4, we can use 1/4 in our proportion instead of 25/100, thus, 1/4 = x/48. Then, cross multiplying, we get 48 = 4x, so x = 12. Then we subtract this discount from the original price yielding 48 – 12 = 36; thus, she paid $36 for the dress.
25% = 25/100
25100=x48
14=x48
(1) * (48) = (4) * (x)
48 = 4x
(48)/4 = (4x)/4
12 = x
The discount is $12.
$48 – $12 = $36
Answer:
f(x) = 2x
Step-by-step explanation:
Remark
What I'm about to do is probably not the best way to do this question, but it is right. The plan is to take out a common factor on the right of f(x - 2) and then derive f(x) from that.
Solution
- f(x - 2) = 2x - 4 Take out a common factor on the right of f(x -2)
- f(x - 2) = 2(x - 2) Now what that means is that in the original equation, wherever you saw an x, you put in x - 2. So the original equation must have been
Check
- f(x) = 2x To check this put x - 2 back in
- f(x - 2) = 2(x -2) Remove the brackets.
- f(x - 2) = 2x - 4 which is what it should be.
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.
