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Reil [10]
4 years ago
12

How many squares will be in the fifth figure

Mathematics
1 answer:
Phantasy [73]4 years ago
5 0

Answer: 30

Step-by-step explanation:

see photo for work

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Could someone show me how to multiply these
mel-nik [20]

Answer:

A.2304

B.5625

C.6776

Step-by-step explanation:

The easiest way to show you is by using calculator. If you can't, try  using Khan Academy if you want.

6 0
3 years ago
Randomly split the messages into a training set D1 (80% of messages) and a testing set D2 (20% of messages). Calculate the testi
Blizzard [7]

Answer:

In the step-by-step

Step-by-step explanation:

This is the code i created using the R software and the packages "caret" and "e0171".

The script was supposed to work in all general cases.

library(caret)

library(e1071)

# Categorical vector

spam <- c("spam","not_spam")

spam_vec <- sample(spam,60, replace = T)

# Supposing two independent variables so that the kappa will be close to 0.

x1 <- rnorm(60)

x2 <- rnorm(60)

# Creating the dataset

data1 <- cbind(spam_vec,x1,x2)

data1 <- as.data.frame(data1)

names(data1) <- make.names(c("spamvec","x","y"))

# Creating the partition

index <- createDataPartition(data1$spamvec,

                            p=0.8, list=FALSE)

training_data <- data1[index,]

testing_data <- data1[-index,]

fitControl <- trainControl(method = "cv",

                         number = 5,

                         savePred = TRUE,

                         classProb = TRUE)

tune.grid <- expand.grid(C = seq(0, 10, .1))

# Scaling the predictors

preProcess_cs <- preProcess(training_data[, -1],

                                 method = c("center", "scale"))

spam_training_cs      <- predict(preProcess_cs, training_data)

spam_testing_cs       <- predict(preProcess_cs, testing_data)

# Training a Naive Bayes to predict binary outcome

Naive_Bayes_Model=naiveBayes(spamvec ~.,

                            data=spam_training_cs,

                            tuneGrid = tune.grid,

                            trControl = fitControl)

# Confusion matrix

prediction <- predict(Naive_Bayes_Model, spam_testing_cs)

confusionMatrix(prediction, spam_testing_cs$spamvec, positive = "spam")

confM <- confusionMatrix(prediction, spam_testing_cs$spamvec, positive = "spam")

accuracy <- confM$overall[1]

accuracy

3 0
3 years ago
Which answer is the explicit rule for the sequence 11, 8.5, 6, 3.5, 1
arsen [322]

I think it's an= 13.5 - 2.5n

7 0
2 years ago
Read 2 more answers
The mean lifetime of a tire is 4242 months with a variance of 4949. If 145145 tires are sampled, what is the probability that th
SVEN [57.7K]

The above question is not correctly written

Complete Question

The mean lifetime of a tire is 42 months with a variance of 49. If 145 tires are sampled, what is the probability that the mean of the sample would be greater than 42.8 months.

Answer:

0.083793

Step-by-step explanation:

We would be using the z score formula.

z score formula = z = (x - μ)/σ/√n

where

x is the raw score = 42.8

μ is the population mean = 42

σ is the population standard deviation =

In the above question, we were given variance = 49

Standard deviation = √Variance

= √49

= 7

n = number of samples = 145

z score = z = (x - μ)/σ/√n

= (42.8 - 42)/ (7/√145)

= 0.8/ 0.581318359

= 1.37618

Approximately to 2 decimal places= 1.38

Using the z score table to find the probability.

P(x ≤ 42.8) = P(z = 1.38) = 0.91621

P(x>42.8) = 1 - P(x<42.8)

1 - 0.91621

= 0.083793

Therefore, the probability that the mean of the sample would be greater than 42.8 is 0.083793

6 0
3 years ago
Please I beg you! Please need help, I can pay you! From PayPal if you help me!!!
Naddika [18.5K]

Answer:

lol how much??????.......

5 0
3 years ago
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