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Lorico [155]
2 years ago
6

Suppose you run a bookstore, and have ratings (1 to 5 stars) of books. Your collaborative filtering algorithm has learned a para

meter vector for user j, and a feature vector for each book. You would like to compute the "training error", meaning the average squared error of your system’s predictions on all the ratings that you have gotten from your users. Which of these are correct ways of doing so (check all that apply)For this problem, let m be the total number of ratings you
have gotten from your users. (Another way of saying this is
that m=∑nmi=1∑nuj=1r(i,j)). [Hint: Two of the four options below are correct.]
a) 1m∑(i,j):r(i,j)=1((θ(j))Tx(i)−r(i,j))2
b) 1m∑nmi=1∑j:r(i,j)=1(∑nk=1(θ(j))kx(i)k−y(i,j))2
c) 1m∑(i,j):r(i,j)=1((θ(j))Tx(i)−y(i,j))2
d) 1m∑nuj=1∑i:r(i,j)=1(∑nk=1(θ(k))jx(k)i−y(i,j))2
Mathematics
1 answer:
LenKa [72]2 years ago
4 0
D because I’m forced to answer this question
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Five more than twice a number is -13 into a two step equation
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2n + 5

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Write 4/14 in simplest form
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2/7

Step-by-step explanation:

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The weight of a cat is normally distributed with a mean of 9 pounds and a standard deviation of 2 pounds. Using the empirical ru
coldgirl [10]

If the value of the z-score is 1. Then the probability that a cat will weigh less than 11 pounds will be 0.84134.

<h3>What is the z-score?</h3>

The z-score is a statistical evaluation of a value's correlation to the mean of a collection of values, expressed in terms of standard deviation.

The z-score is given as

z = (x - μ) / σ

Where μ is the mean, σ is the standard deviation, and x is the sample.

The weight of a cat is normally distributed with a mean of 9 pounds and a standard deviation of 2 pounds.

Then the probability that a cat will weigh less than 11 pounds will be

The value of z-score will be

z = (11 – 9) / 2

z = 1

Then the probability will be

P(x < 11) = P(z < 1)

P(x < 11) = 0.84134

Thus, the probability that a cat will weigh less than 11 pounds will be 0.84134.

More about the z-score link is given below.

brainly.com/question/15016913

#SPJ1

4 0
2 years ago
The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population o
blondinia [14]

The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

According to the statement

we have to find that the standard deviation, mean and the intervals from the given data.

So, According to the given data from the method of tree ring dating

The value of mean is

x-bar = (1271 + 1208 + 1229 + 1299 + 1268 + 1316 + 1275 + 1317 + 1275) / 9 = x-bar = 1273 AD

And now we find standard deviation :

s = √∑(xi - x-bar) / (N - 1)

∑(xi - x-bar)^2 = (1271 - 1273)2 + (1208 - 1273)2 + (1229 - 1273)2 + ... + (1275 - 1273)2

∑(xi - x-bar)^2 = (-2)2 + (-65)2 + (-44)2 + ... + (2)2

∑(xi - x-bar)^2 = 4 + 4225 + 1936 + 676 + 25 + 1849 + 4 1936 + 4

∑(xi - x-bar)^2 = 10,659

Now,

s^2 = 10659/8 = 1332

s = 37 years

So, standard deviation is 37 years.

We need the t-distribution table since the standard deviation is unknown.  Therefore, our degrees of freedom is 9 - 1 = 8 and the critical value is 1.860.  Set up the confidence interval for the mean:

[x-bar ± t*(s/√n)] = [1273 ± 1.860*(37/√9)]

[x-bar ± t*(s/√n)] = [1250,1296]

So, The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

Learn more about method of tree ring dating here

brainly.com/question/15107034

Disclaimer: This question was incomplete. Please find the full content below.

Question:

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.

For more data please see the image below.

#SPJ4

4 0
2 years ago
-3(2x^2+ax+b)=-6x+12x-15 what is the value of a and b
bonufazy [111]
B= -2x+5-2x^2-ax
A= -2x+5-2x^2-b/x 
x=0

5 0
3 years ago
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