What is the Constant of Proportionality for this table?

Daniel makes 100 dollars each week. He worked for a weeks this summer.

Digiron [165]

Answer:

the number of weeks is unknown the A represents the number of weeks so the answer is 100a

7 0

3 years ago

An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos

Schach [20]

Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.78, n = 675$

So

$\mu = E(X) = np = 675*0.78 = 526.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76$

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when $X = 0.8*675 = 540$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{540 - 526.5}{10.76}$

$Z = 1.25$

$Z = 1.25$ has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

8 0

3 years ago

Find the value of a. 19 in a 99 in Perimeter 238 inches

Sonja [21]

Answer:

the answer is 60 inches

Step-by-step explanation:

you have to subtract the perimeter and the numbers that are shown after you add them

238-118= 120 then you have to divide that in half wich =60.

so (A)= to 60 inches.

6 0

3 years ago

How do i factor 2x^2+11x+5

ira [324]

For simple integer factors as this one has, you want to find two values for the quadratic in the form ax^2+bx+c. Let the two values be j and k. These two values must satisfy two conditions.

jk=ac=10 and j+k=b=11, so j and k must be 1 and 10.

Now replace bx with jx and kx...

2x^2+x+10x+5 now factor 1st and 2nd pair of terms.

x(2x+1)+5(2x+1)

(x+5)(2x+1)

7 0

3 years ago

Read 2 more answers

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 $\left( {a + b} \right)(a+b)$ .

The classification error rate for the records those are truly fraudulent is obtained as shown below:

The classification matrix is, shown above and in the attachment

The error rate for truly fraudulent is,

$\begin{array}{c}\\FP = \frac{b}{{a + b}}\\\\ = \frac{{58}}{{30 + 58}}\\\\ = \frac{{58}}{{88}}\\\\ = 0.6591\\\end{array}$

The percentage is, $0.6591 \times 100 = 65.91$

(b)

The classification error rate for records that are truly non-fraudulent is obtained by taking the ratio of d and $\left( {c + d} \right)(c+d)$ .

The classification error rate for records that are truly non-fraudulent is obtained as shown below:

The classification matrix is, shown in the attachment

The error rate for truly non-fraudulent is,

$\begin{array}{c}\\TP = \frac{d}{{c + d}}\\\\ = \frac{{920}}{{32 + 920}}\\\\ = \frac{{920}}{{952}}\\\\ = 0.9664\\\end{array}$

The percentage is, $0.9664 \times 100 = 96.64$

8 0

3 years ago