Moving average method - Class IX

From Permutation and Combination,

$\cfrac{A+B+C}{3}=45$ $kg$

Let his present age be x
According to problem x + 12 = 4 (x - 12)
-3x = - 48 - 12
3x = 60
x = 20 years

$\Rightarrow$  Moving-averages : $Smooth-out,the,time,series$.

These statements are not enough to find out his English marks. 
So Ans = D

Average speed is total distance divided by total time. Total distance is 160 + 160 = 320 km. Time 1 = 160/64 = 2.5 hours. Time 2 = 160/80 = 2 hours. Total time = 4.5 hours. Average speed = 320 / 4.5 = 71.11 km/h.

Only Bira and his elder daughter can be called unemployed as per the given situation.

Qualitative forecast is an estimation methodology which uses expert judgment, rather than numerical analysis. It predicts the direction, but not the magnitude of change in variable.

$\displaystyle \frac{1}{20} + \frac{1}{30} = \frac{1}{x}$
$\displaystyle \frac{5}{60} = \frac{1}{x}     \Rightarrow 12 min$

The first step in time series analysis is to plot the data on a graph.

In moving average method, we cannot find the trend value of some:between extreme periods.

$\Rightarrow$  The method of least squares dictates that we choose a regression line where the sum of the square of deviations of the points from the line is $:Minimum$

$\Rightarrow$  In simple linear regression, the number of unknown constants are $:Two$

Both the regression coefficients $(b_{xy},b_{yx})$ must have the same sign. i.e., if one of them is positive other should positive or if one of them is negative other should be negative.

The regression model gives the relation between two or more variables.

In the exponential growth model y = x0(1+r)^x, x0 represents the initial value when x = 0. Since x is the number of years after birth, x = 0 corresponds to the year of birth.

A linear regression line has an equation of the form $Y=a+bX$

Given that observed value is $50$ and

Let the actual values be $y_1,y_2,...,y_n$

Let the actual values be $y_1,y_2,...,y_n$

The linear equation be y=a+bx

When the regression line is linear $(y = ax + b)$ the regression coefficient is the constant $(a)$ that represents the rate of change of one variable $(y)$ as a function of changes in the other $(x)$ i.e. it is the slope of the regression line. Hence, it measures relationship between 2 variables but does not always account for exact change in value of 1 variable due to unit change in another variable due to non-zero value of $(b)$.

Using the linear regression formula for x on y: x - mean_x = b_xy * (y - mean_y). Mean_x = 125/25 = 5. Mean_y = 100/25 = 4. Calculating b_xy and solving for x when y=5 yields approximately 4.591.

$\bar x=18$, $\bar y=100$, $\sigma_x=14$, $\sigma_y=20$ and $r=0.8$
Regression line of y on x will be $y-\bar y=r\dfrac{\sigma_y}{\sigma_x}(x-\bar x)$
Subsitute all the abpve value in an equation of regression line, we get
$y-100=\dfrac{0.8\times 20}{14}(x-18)$
$y-100=1.143(x-18)$
$y=1.143x+79.43$
Now, Subsitute the value of x=70, we get
$y=1.143\times 70+79.43=159.43$

Let the equation be $ax+by+c=0$

Let us assume that $7x-3y-18=0$ is the regression equation of $y$ on $x$.

$\bar x=20$, $\bar y=90$, $\sigma_x=5$, $\sigma_y=12$ and $r=0.8$
Regression line of y on x will be $y-\bar y=r\dfrac{\sigma_y}{\sigma_x}(x-\bar x)$
Subsitute all the abpve value in an equation of regression line, we get
$y-90=\dfrac{0.8\times 12}{5}(x-20)$
$y-90=1.92(x-20)$
$y=1.92x+90-38.4$
Now, Subsitute the value of x=15, we get
$y=1.92\times 15+51.6=80.4$

$\sum x=30$
$\sum y=30$
$\sum x^2=190$
$\sum x\times y=192$
So, $\bar x=\dfrac{\sum x}{n}=\dfrac{30}{5}=6$
$\bar y=\dfrac{\sum y}{n}=\dfrac{30}{5}=6$
$b_{yx}=\dfrac{\sum xy-\dfrac{\sum x \times \sum y}{n}}{\sum x^2-\dfrac{(\sum
x)^2}{n}}=\dfrac{192-180}{190-180}=\dfrac{12}{10}=1.2$
Regression line of y on x will be $y-\bar y=b_{yx}(x-\bar x)$
$y-6=1.2(x-6)$
$y=1.2x-7.2+6$
$y=1.2x-1.2$

Since the two lines of regression interest at the point $(\bar{X}, \bar{Y} )$

Given two lines $x+2y-5=0, x+3y-8=0$.