can someone help me I’m stuck on this and can’t figure it out

Answer:

51.5?

Step-by-step explanation:

I think it's just length times width divided by 2

The area of the kite ABCD is 51.56 square yards.

A kite's area is the amount of space enclosed or surrounded by a kite in a two-dimensional plane.

A kite's area equals half the product of its diagonal lengths. The formula for calculating a kite's area is:

Area = 1/2 × d₁ × d₂

Where d₁ and d₂ are long and short diagonals of a kite.

Here, AC = 12.5 yards and BD = 8.25 yards

Substitute the values of d₁ = 12.5 yards and d₂ in the formula of the kite's area:

The area of the kite = 1/2 × 12.5 × 8.25

The area of the kite = 51.5625

The area of the kite ≈ 51.56 square yards

Thus, the area of the kite ABCD is 51.56 square yards.

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Answer: 0.1562 I believe

Answer:

27.5

Step-by-step explanation:

pythagorean theorem. a^2+b^2=c^2.

so 12^2+b^2=30^2

meaning 144+b^2=900

900-144=756

square root 756

b=27.5

Answer:

30

Step-by-step explanation:

15 x 2

After testing hypothesis with alternative-hypothesis is not equal to 30%,  the p-value is 0.278.

To test the hypothesis that proportion of unemployed people in Karachi city is not equal to 30%, we perform a two-tailed test using binomial distribution.

The null-hypothesis (H₀) is that the proportion of unemployed people is 30% (p = 0.30), and the alternative-hypothesis (H₁) is that the proportion is not equal to 30% (p ≠ 0.30),

We have a sample of 100 people, and 35 of them are unemployed. we will calculate "test-statistic" "z-score" and use it to find p-value,

The "test-statistic" formula for "two-tailed" test is :

z = (p' - p₀)/√((p₀ × (1 - p₀))/n),

where p' = sample proportion, p₀ = hypothesized-proportion under the null-hypothesis, and n = sample-size,

In this case, p' = 35/100 = 0.35, p₀ = 0.30, and n = 100,

Calculating the test statistic:

z = (0.35 - 0.30)/√((0.30 × (1 - 0.30)) / 100)

= 0.05/√((0.30 * 0.70) / 100)

≈ 0.05/√(0.21 / 100)

≈ 0.05/√(0.0021)

≈ 0.05/0.0458258

≈ 1.0905

To find the p-value, we calculate probability of obtaining "test-statistic" as extreme as 1.0905 in a two-tailed test.

We know that the cumulative-probability (area) to left of 1.0905 is approximately 0.861.

Since this is a two-tailed test, we double this probability to get "p-value":

So, p-value = 2 × (1 - 0.861),

= 2 × 0.139,

= 0.278

Therefore, the p-value for hypothesis test is 0.278.

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Answer:

c

Step-by-step explanation:

Answer:

c

Step-by-step explanation:

Answer:

2

6

9

Step-by-step explanation:

Because if you pay attention to our teacher he told us how to respond it!

Answer:

[tex]f(8)=74[/tex]

Step-by-step explanation:

[tex]f(8)=8^2+10=64+10=74[/tex]

Answer:

82944 in^3

Step-by-step explanation:

Volume = length x breadth x height

Before volume can be calculated, feet and yards needs to be converted to inches

1 yard = 36 inches

2 x 36 = 72 inches

1 foot = 12 inches

2 x 12 = 24 inches

Volume = 72 x 48 x 24 = 82944 in^3

a right angle so 90 degrees

Answer:

x=18

y=27

explanation: 4 times 3 is 12

Answer: 5 people

Step-by-step explanation:

The easiest way to do this is to set up a proportion. 6/12=x/10. To solve, cross-multiply and divide. Multiply 6 and 10 (6*10=60) and divide by 12 (60/12=5).

You can also find that 6/12 is equal to 1/2 by dividing each side by 6. Then multiply 10 by 1/2 (or divide 10 by 2) to get the final answer of 5 people.

Main answer: RP* = RP.

Supporting answer:

RP* is defined as the set of languages L for which there exists a polynomial-time Turing machine M that, on input w, halts with probability at least 1/2 if w is not in L, and halts with probability at least 1 - 2-l|w| if w is in L. This definition is almost identical to the definition of RP, except that the probability of error is reduced from 1/2 to 2-l|w|. However, RP* and RP are equivalent.

To see that RP* is a subset of RP, note that if a language L is in RP*, then there exists a polynomial-time Turing machine M and a polynomial p such that (i) if we L, then Prųe{0,1 }p(w) [M accepts (w,t)]21-(1/2)lwl and (ii) if w & L, then Prge{0,1}P(w) [M rejects (w,t)] = 1. This means that M can be used as a randomized algorithm for L with error probability at most 1/2. Therefore, L is in RP.

To see that RP is a subset of RP*, note that if a language L is in RP, then there exists a polynomial-time Turing machine M and a constant c such that (i) if we L, then Prųe{0,1}c|w|[M accepts (w,t)] >= 1/2 and (ii) if w & L, then Prge{0,1}P(w) [M rejects (w,t)] < 1/2. By setting p(n) = c * 2n, we obtain a Turing machine M' that satisfies the conditions of RP*. Therefore, L is in RP*.

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Answer:

2.4

Step-by-step explanation:

3 / 28 = 0.12

(we do this to see how many miles he ran in a minute)

0.12 x 20 = 2.4

(then we multiply how many miles he can run in 20 minutes)

Given: For a standard normal distribution, we need to find the probability between P(-2.46 < z < 2.82). It is found that P(-2.46 < z < 2.82) = 0.9905.

Explanation: Given, P(-2.46 < z < 2.82)

The standard normal distribution has a mean of μ=0 and a standard deviation of σ=1. It is called the standard normal distribution, because it is the normal distribution where z-scores correspond to the number of standard deviations above or below the mean.

A z-score tells us how many standard deviations a value is from the mean.

A positive z-score indicates a value above the mean, while a negative z-score indicates a value below the mean.

To find the probability of P(-2.46 < z < 2.82), we need to find the area under the standard normal distribution curve between -2.46 and 2.82.

To find this probability, we can use a standard normal distribution table or a calculator that has a normal distribution function.

Using a standard normal distribution table, we can find the area to the left of z=2.82 and the area to the left of z=-2.46 and then subtract the two values to find the area between these z-scores.

The area to the left of z=2.82 is 0.9974, and the area to the left of z=-2.46 is 0.0069.

Therefore, the area between these z-scores is:

P(-2.46 < z < 2.82) = 0.9974 - 0.0069

= 0.9905

Therefore, P(-2.46 < z < 2.82) = 0.9905.

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Answer:

- 14

Step-by-step explanation:

c² - d² ← is a difference of squares and factors as

= (c + d)(c - d) ← substitute values for factors

= 7 × - 2

= - 14

If an analyst wants to estimate the mean for an entire population (μ), the estimate would be more accurate if the analyst computed the sample mean.

If an analyst wants to estimate the mean for an entire population (μ), the estimate would be more accurate if the analyst computed the sample mean.

What is a population?In statistics, a population is a complete set of events that a statistician desires to investigate. A population is a collection of individuals, items, or data points that have a specific attribute of interest to the analyst.

What is an estimate? An estimate is an approximation of an unknown quantity that is dependent on imperfect or incomplete information. In statistics, an estimate is a projection of a population parameter dependent on data collected from a sample. An estimate is a numerical value generated from a statistical formula that is intended to provide an approximate value for an unknown population parameter.

What is an analyst?An analyst is an individual who examines a company or business's financial and business data to evaluate their health and determine their future development.

What does it mean to estimate the mean for an entire population?In statistics, estimating the mean for an entire population entails using a sample of data to calculate an approximate value of the mean for the entire population. The mean is the numerical value that provides information about the data set's central tendency. The analyst should choose a representative sample of the population to ensure the estimate is accurate.

What is the best way to estimate the mean of the entire population?

The estimate would be more accurate if the analyst computed the sample mean. The sample mean is an estimate of the population mean, denoted as μ. Sample mean is the average of the sampled data, and it is computed as follows;$$\overline{x}=\frac{\sum_{i=1}^{n}x_i}{n}$$

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If an analyst wants to estimate the mean for an entire population (μ), the estimate would be more accurate if the analyst computed the mean of a larger random sample.

A random sample is a method of selecting a subset from the entire population. It should be such that every member of the population has an equal chance of being chosen.

The sample should be sufficiently large and representative of the population as a whole to make reasonable inferences regarding the population. The mean value computed from a sample is used as an estimate of the true population mean. The sample mean is a random variable that can fluctuate from one sample to the next, depending on the sample that is taken. It has a standard deviation, called the standard error of the mean, that can be calculated using the following formula:standard error of the mean = standard deviation of the population / square root of the sample size.The larger the sample size, the smaller the standard error of the mean, and hence the more accurate the estimate of the population mean will be.

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The probability that their mean systolic blood pressure is between 119 and 122 is 0.0807.

The given distribution is normal.

So, the formula for the standardized random variable, z can be used.

Here,Mean of the given distribution, μ = 114.8

Standard deviation of the given distribution, σ = 13.1

Number of women aged 18-24 randomly selected, n = 23

Let X be the mean systolic blood pressure of 23 randomly selected women aged 18-24.

P(X is between 119 and 122) = P((X-μ)/σ is between (119-μ)/(σ/√n) and (122-μ)/(σ/√n))

= P((X-μ)/σ is between (119-114.8)/(13.1/√23) and (122-114.8)/(13.1/√23))

= P((X-μ)/σ is between 1.35 and 2.45)

Using standard normal distribution table,P(1.35 < z < 2.45)= P(z < 2.45) - P(z < 1.35)≈ 0.9922 - 0.9115= 0.0807

Thus, the probability that the mean systolic blood pressure of the randomly selected 23 women aged 18-24 is between 119 and 122 is approximately equal to 0.0807.

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Using the Excel Regression tool on the Weddings dataset with wedding cost as the dependent variable and attendance as the independent variable, the key regression results, hypothesis tests, and confidence intervals should be interpreted.

After applying the Excel Regression tool with wedding cost as the dependent variable and attendance as the independent variable, the key regression results should be examined. These results typically include coefficients, standard errors, t-values, p-values, and confidence intervals. The coefficient for the independent variable (attendance) represents the estimated change in the wedding cost for each unit increase in attendance. Hypothesis tests and confidence intervals help assess the statistical significance and precision of the estimated coefficients. The t-value indicates the strength of evidence against the null hypothesis of no relationship between the variables, and the p-value indicates the probability of observing the estimated coefficient under the null hypothesis. Lower p-values suggest stronger evidence against the null hypothesis. To validate the assumptions of the regression analysis, the residuals (the differences between the observed and predicted values) should be analyzed. Residual plots can be examined to check for patterns or deviations from assumptions, such as linearity, independence, and constant variance. Outliers can be identified by examining the standardized residuals. Values with high absolute values indicate potential outliers. To estimate the budget for a wedding with 175 guests, the regression model can be used to predict the corresponding cost. The attendance value of 175 can be plugged into the regression equation, and the predicted wedding cost can be obtained.

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If a pair of dice is rolled, and the number that appears uppermost on each die is observed, the probability of the sum of the numbers being either 7 or 11 is 2/9.

To find the probability of the sum of the numbers rolled on a pair of dice being either 7 or 11, we need to determine the number of favorable outcomes and the total number of possible outcomes.

There are six possible outcomes for each die, ranging from 1 to 6. Since we are rolling two dice, the total number of possible outcomes is 6 multiplied by 6, which is 36.

To calculate the number of favorable outcomes, we need to determine the combinations that result in a sum of either 7 or 11.

For the sum of 7, there are six possible combinations: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1).

For the sum of 11, there are two possible combinations: (5, 6) and (6, 5).

Therefore, the number of favorable outcomes is 6 + 2 = 8.

The probability of the sum of the numbers being either 7 or 11 is given by the ratio of favorable outcomes to the total number of outcomes:

P(sum is 7 or 11) = favorable outcomes / total outcomes = 8/36 = 2/9.

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Complete question is:

A pair of dice is rolled, and the number that appears uppermost on each die is observed. Refer to this experiment and find the probability of the given event. (Enter your answer as a fraction.)

The sum of the numbers is either 7 or 11.

a) The estimated model relating annual salary to firm sales and market value, in equation form, is: Salary = 4.6209 + 0.1621 * log(sales) + 0.1067 * log(mktval), where log denotes the natural logarithm.

b) Calculating this expression will give us the expected salary according to the model. If the expected salary is higher than $500,000, then your friend would be asking too much.

a) The estimated model relating annual salary to firm sales and market value, in equation form, is:

Salary = 4.6209 + 0.1621 * log(sales) + 0.1067 * log(mktval)

where log denotes the natural logarithm.

b) To determine if your friend would be asking too much for an annual salary of $500,000, we need to plug the values of firm sales and market value into the model and calculate the expected salary.

Using the given values:

- Firm sales (sales) = $5,000,000

- Market value (mktval) = $20,000,000

We first need to take the logarithm of the sales and market value:

log(sales) = log(5,000,000)

log(mktval) = log(20,000,000)

Then, we can substitute these values into the equation:

Expected Salary = 4.6209 + 0.1621 * log(5,000,000) + 0.1067 * log(20,000,000)

Calculating this expression will give us the expected salary according to the model. If the expected salary is higher than $500,000, then your friend would be asking too much.

Note: Make sure to use the natural logarithm (ln) in the calculations.

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The value of the given integral is 2π [sin(20e) - a² sin(20e) - (a²/2) sin(20e) + (a²/2) (1/2) [sin(22e)/22 + sin(18e)/18] + C]

To evaluate the given expression:

∫∫ cos(20e) (1 - 2a cos(e) + a²) de dθ

We need to integrate with respect to both e and θ.

First, let's integrate with respect to e:

∫ cos(20e) (1 - 2a cos(e) + a²) de

Using the power reduction formula for cosine, we can rewrite the integrand:

= ∫ cos(20e) (1 - a² + a² cos²(e) - 2a cos(e)) de

= ∫ cos(20e) (1 - a² - a² sin²(e)) de

Now, we can integrate term by term:

= ∫ cos(20e) de - a² ∫ cos(20e) sin²(e) de - a² ∫ cos(20e) de

The first and third integrals are straightforward to evaluate:

= ∫ cos(20e) de - a² ∫ cos(20e) de - a² ∫ cos(20e) sin²(e) de

= sin(20e) - a² sin(20e) - a² ∫ cos(20e) sin²(e) de

Now, let's evaluate the remaining integral with respect to e:

= sin(20e) - a² sin(20e) - a² ∫ cos(20e) sin²(e) de

Using the trigonometric identity sin²(e) = (1 - cos(2e))/2, we can simplify the integrand:

= sin(20e) - a² sin(20e) - a² ∫ cos(20e) ((1 - cos(2e))/2) de

= sin(20e) - a² sin(20e) - (a²/2) ∫ cos(20e) - cos(20e) cos(2e) de

= sin(20e) - a² sin(20e) - (a²/2) ∫ cos(20e) de + (a²/2) ∫ cos(20e) cos(2e) de

Integrating the first term gives:

= sin(20e) - a² sin(20e) - (a²/2) ∫ cos(20e) de + (a²/2) ∫ cos(20e) cos(2e) de

= sin(20e) - a² sin(20e) - (a²/2) sin(20e) + (a²/2) ∫ cos(20e) cos(2e) de

Next, let's evaluate the remaining integral with respect to e:

= sin(20e) - a² sin(20e) - (a²/2) sin(20e) + (a²/2) ∫ cos(20e) cos(2e) de

Using the trigonometric identity cos(a) cos(b) = (1/2) [cos(a + b) + cos(a - b)], we can simplify the integrand:

= sin(20e) - a² sin(20e) - (a²/2) sin(20e) + (a²/2) ∫ (1/2) [cos(20e + 2e) + cos(20e - 2e)] de

= sin(20e) - a² sin(20e) - (a²/2) sin(20e) + (a²/2) (1/2) [∫ cos(22e) de + ∫ cos(18e) de]

= sin(20e) - a² sin(20e) - (a²/2) sin(20e) + (a²/2) (1/2) [sin(22e)/22 + sin(18e)/18] + C

where C is the constant of integration.

Now, we have the integral with respect to e evaluated. To find the final result, we need to integrate with respect to θ. However, we don't have any dependence on θ in the original expression. Therefore, we can simply multiply the result obtained above by 2π, as we're integrating over the entire range of θ.

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Complete question:

The number of different simple random samples of size 5 that can be obtained from a population of size 35 is 324,632.

To calculate the number of different simple random samples of size 5 that can be obtained from a population of size 35, we can use the combination formula. The formula for combination is given by:

C(n, r) = n! / (r! * (n-r)!)

where n is the population size and r is the sample size.

In this case, we have n = 35 (population size) and r = 5 (sample size). Plugging in these values into the formula:

C(35, 5) = 35! / (5! * (35-5)!)

Simplifying this expression:

C(35, 5) = 35! / (5! * 30!)

Calculating the factorial values:

C(35, 5) = 35 * 34 * 33 * 32 * 31 / (5 * 4 * 3 * 2 * 1)

C(35, 5) = 324,632

Therefore, the number of different simple random samples of size 5 that can be obtained from a population of size 35 is 324,632.

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Answer: I think A

Step-by-step explanation:

A) The scatterplot of the data is as follows:What can we interpret from the scatterplot of the given data?We can observe that the higher the number of risk factors, the lower the IQ of the children. This fact is supported by the trend line that shows a negative correlation. This is what the plot is telling us.A regression equation is given by the formula: $y=\beta_0+\beta_1x$Where, $y$ is the IQ of the child, $x$ is the number of risk factors, $\beta_0$ is the y-intercept and $\beta_1$ is the slope of the line.Both $\beta_0$ and $\beta_1$ can be calculated using the following formulae: $$\beta_1=r_{xy}\frac{s_y}{s_x}$$$$\beta_0=\bar{y}-\beta_1\bar{x}$$Where, $r_{xy}$ is the correlation coefficient between $x$ and $y$, $s_x$ and $s_y$ are the standard deviations of $x$ and $y$ respectively, and $\bar{x}$ and $\bar{y}$ are the means of $x$ and $y$ respectively.From the scatterplot, we can observe that $r_{xy}$ is negative. Therefore, there exists a negative correlation between $x$ and $y$.Let us calculate the values of $\beta_1$, $\beta_0$ using the above formulae.$$r_{xy}=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sqrt{\sum(x-\bar{x})^2\sum(y-\bar{y})^2}}$$$$r_{xy}=\frac{(3)(112)+(6)(82)+(0)(105)+(5)(102)+(1)(115)+(1)(101)+(5)(94)+(4)(89)+(3)(98)+(3)(109)+(4)(91)+(2)(107)}{\sqrt{(3^2+6^2+0^2+5^2+1^2+1^2+5^2+4^2+3^2+3^2+4^2+2^2)(112^2+82^2+105^2+102^2+115^2+101^2+94^2+89^2+98^2+109^2+91^2+107^2)}}$$$$r_{xy}=-0.835$$Now, let's calculate the standard deviations of $x$ and $y$:$$s_x=\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$$$$s_y=\sqrt{\frac{\sum(y-\bar{y})^2}{n-1}}$$$$s_x=\sqrt{\frac{(3-2)^2+(6-2)^2+(0-2)^2+(5-2)^2+(1-2)^2+(1-2)^2+(5-2)^2+(4-2)^2+(3-2)^2+(3-2)^2+(4-2)^2+(2-2)^2}{12-1}}$$$$s_x=\sqrt{\frac{44}{11}}=2$$$$s_y=\sqrt{\frac{\sum(y-\bar{y})^2}{n-1}}$$$$s_y=\sqrt{\frac{(112-99.17)^2+(82-99.17)^2+(105-99.17)^2+(102-99.17)^2+(115-99.17)^2+(101-99.17)^2+(94-99.17)^2+(89-99.17)^2+(98-99.17)^2+(109-99.17)^2+(91-99.17)^2+(107-99.17)^2}{12-1}}$$$$s_y=\sqrt{\frac{5626.1}{11}}=9.025$$Finally, let's calculate $\beta_1$ and $\beta_0$:$$\beta_1=r_{xy}\frac{s_y}{s_x}$$$$\beta_1=-0.835\frac{9.025}{2}=-3.762$$$$\beta_0=\bar{y}-\beta_1\bar{x}$$$$\beta_0=99.17-(-3.762)(3.08)=112.91$$Therefore, the regression equation is $y=112.91-3.762x$.Here, $\beta_0=112.91$ represents the expected IQ of a child with zero risk factors, while $\beta_1=-3.762$ represents the decrease in IQ per risk factor.B) James has 5 risk factors. Substituting $x=5$ in the regression equation, we get:$$y=112.91-3.762x$$$$y=112.91-3.762(5)$$$$y=93.13$$Therefore, the IQ predicted for James is $93.13$.Similarly, Elizabeth has 2 risk factors. Substituting $x=2$ in the regression equation, we get:$$y=112.91-3.762x$$$$y=112.91-3.762(2)$$$$y=105.39$$Therefore, the IQ predicted for Elizabeth is $105.39$.

The required expression is: y' = (1/2)(1-y)/(1+y)

Given [1 + y]² - x + y = 4 10. We need to determine an expression for dy/dx = y'.

Simplification of the given expression:

[1 + y]² + y - 4 10 = x

Differentiating w.r.t x by using the chain rule, we get:

(2[1 + y])*(dy/dx) + dy/dx + 1 = 0

(dy/dx)[2(1 + y) + 1] = - 1 - [1 + y]²

(dy/dx) = [- 1 - (1 + y)²]/[2(1 + y)]

The given expression is [1+y]²-x+y=4 10. We need to determine an expression for dy/d x = y'.

Differentiating the given equation with respect to x, we get:

2(1+y).dy/dx - 1 + dy/dx = 0

dy/dx(2+2y) = 1 - y(2+dy/dx)

dy/dx(2+2y) = (1-y)(2+dy/dx)

dy/dx = (1-y)/(2+2y)

dy/dx = (1/2)(1-y)/(1+y)

Hence, the required expression is: y' = (1/2)(1-y)/(1+y)

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Answer:

f(1) = 14000

f(n) = f(n - 1) · 0.9 , for n ≥ 2

The next term in the sequence is 10206 .

Step-by-step explanation:

First we try to find the type of sequence it is.

Is there a common difference?

12600 - 14000 = -1400

11340 - 12600 = -1260

There is no common difference, so it is not an arithmetic sequence.

12600/14000 = 0.9

11340/12600 = 0.9

There is a common ratio, so this is a geometric sequence.

Each term is 0.9 times the previous term.

The next term is:

11340 * 0.9 = 10206

Answer:

f(1) = 14000

f(n) = f(n - 1) · 0.9 , for n ≥ 2

The next term in the sequence is 10206 .

(a) The integral of w(x) over its entire domain is zero, it cannot be equal to 1. This means that the given function does not satisfy the normalization property and hence cannot be a probability density function.

To show that the given probability density function is a valid one, we need to verify that it satisfies the two properties of a probability density function:

1. Non-negativity:

For 0 <= x <= 9, c(9-x) is always non-negative, since c is a positive constant and (9-x) is also non-negative in this range. For any other value of x, w(x) is zero. Hence, w(x) is non-negative for all x.

2. Normalization:

[tex]a[0,9] w(x) dx = a[0,9] c(9-x) dx[/tex]

= [tex]c a[0,9] (9-x) dx[/tex]

= [tex]c [(9x - (x^2)/2)] [from 0 to 9][/tex]

= [tex]c [(81/2) - (81/2)][/tex]

= [tex]c (0)[/tex]

=[tex]0[/tex]

(b) The given probability density function does not have a valid normalization constant and hence does not represent a valid probability distribution.

To find the mean life time of a component, we need to calculate the expected value of X using the formula:

[tex]E(X) = a[a,b] x (w(x) dx)[/tex]

where a and b are the lower and upper bounds of the domain respectively.

In this case, we have:

a = 0 and b = 9

w(x) = c(9-x)

Hence,

[tex]E(X) = a[0,9] x*c(9-x) dx[/tex]

= [tex]c a[0,9] (9x - x^2) dx[/tex]

= [tex]c [(81 x^2/2) - (x^3/3)] [from 0 to 9][/tex]

=[tex]c [(6561/2) - (729/3)][/tex]

= [tex]c (2958/3)[/tex]

To find the value of c, we can use the normalization property:

[tex]a[0,9] w(x) dx = 1[/tex]

[tex]a[0,9] c(9-x) dx = 1[/tex]

[tex]c a[0,9] (9-x) dx = 1[/tex]

[tex]c [(81/2) - (81/2)] = 1[/tex]

[tex]c * 0 = 1[/tex]

This is not possible, since c cannot be infinite.

To know more about probability density function refer here:

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