please tel me the answer I’m struggling <br> Also I’ll make you brainlist

Mr. Coletti made a large bowl of chocolate pudding. The pudding weighed 4 1/3 pounds. If he served equal portions of 1/6 pound t

snow_tiger [21]

$\frac{4\frac{1}{3}}{\frac{1}{6}}\\4\frac{1}{3}*6\\\frac{13}{3}*6\\26$

D. 26 portions.

4 0

3 years ago

Cereal. A dietician obtains the amounts of sugar (incentigrams) from 100 centigrams (or 1 gram) in each of 10 differentcereals,

nadezda [96]

Answer:

Range is 44

mean is 29.7

variance is 289.56

standard deviation is 17.016

Step-by-step explanation:

The data given as follows

3, 24, 30, 47, 43, 7, 47,13, 44, 39

Range = maximum - minimum

= 47 -3 = 44

mean = $\frac{3+24+30+47+43+7+47+13+44+39}{10}$ = 297/10 = 29.7

variance = $\frac{sum xi^2 }{N} - mean^2 }$

= 289.56

SD = $\sqrt{variance}$ = 17.016

The standard deviation of the given data is not a good measure as the range and standard deviation are far apart. The standard deviation gives how the given data is dispersed from the mean or expected value, while the range gives a rough idea of variability of the data.

3 0

3 years ago

A company produces computer batteries. Everyday, the company produces exactly 500 batteries. To keep track of how many batteries

valkas [14]

Using the binomial distribution, it is found that about 75 batteries each day are defective.

For each battery, there are only two possible outcomes, either it is defective, or it is not. The probability of a battery being defective is independent of any other battery, hence the <em>binomial distribution</em> is used to solve this question.

<h3>What is the binomial probability distribution?</h3>

It is the probability of exactly <u>x successes on n repeated trials, with p probability</u> of a success on each trial.

The expected value of the binomial distribution is:

$E(X) = np$

In this problem:

3 out of 20 batteries are defective, hence p = 3/20 = 0.15.

Each day, 500 batteries are produced, hence n = 500.

Then, the expected number of defective batteries in a day is given by:

E(X) = np = 500(0.15) = 75.

More can be learned about the binomial distribution at brainly.com/question/14424710

5 0

3 years ago

A data mining routine has been applied to a transaction dataset and has classified 88 records as fraudulent (30 correctly so) an

chubhunter [2.5K]

Answer:

The classification matrix is attached below

Part a

The classification error rate for the records those are truly fraudulent is 65.91%.

Part b

The classification error rate for records that are truly non-fraudulent is 96.64%

Step-by-step explanation:

The classification matrix is obtained as shown below:

The transaction dataset has 30 fraudulent correctly classified records out of 88 records, that is, 30 records are correctly predicted given that an instance is negative.

Also, there would be 88 - 30 = 58 non-fraudulent incorrectly classified records, that is, 58 records are incorrectly predicted given that an instance is positive.

The transaction dataset has 920 non-fraudulent correctly classified records out of 952 records, that is, 920 records are correctly predicted given that an instance is positive.

Also, there would be 952 - 920 = 32 fraudulent incorrectly classified records, that is, 32 records incorrectly predicted given that an instance is negative.

That is,

Predicted value

Active value Fraudulent Non-fraudulent

Fraudlent 30 58

non-fraudulent 32 920

The classification matrix is obtained by using the information related to the transaction data, which is classified into fraudulent records and non-fraudulent records.

The error rate is obtained as shown below:

The error rate is obtained by taking the ratio of $\left( {b + c} \right)(b+c)$ and the total number of records.

The classification matrix is, shown above

The total number of records is, $30 + 58 + 32 + 920 = 1,040$

The error rate is,

$\begin{array}{c}\\{\rm{Error}}\,{\rm{rate}} = \frac{{b + c}}{{{\rm{Total}}}}\\\\ = \frac{{58 + 32}}{{1,040}}\\\\ = \frac{{90}}{{1,040}}\\\\ = 0.0865\\\end{array}$

The percentage is $0.0865 \times 100 = 8.65$

(a)

The classification error rate for the records those are truly fraudulent is obtained by taking the rate ratio of b and $\left( {a + b} \right)(a+b)$ .