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Solve the equation for x: x = 5/25 - 7 + 60. Also, perform regression analysis on the given data.

Mathematics

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3 parts
1
53.2
2

y=0.14195+0.00286xy = 0.14195 + 0.00286x

3
0.19919

Step-by-step solution

9 steps

Here's the solution to your questions:

Part 1· 3 steps

Solve the equation for $س$ (x)

  1. Write down the equation.

    س=525−7+60س = \frac{5}{25} - 7 + 60
  2. Simplify the fraction.

    س=0.2−7+60س = 0.2 - 7 + 60
  3. Perform the subtraction and addition.

    س=−6.8+60س = -6.8 + 60 س=53.2س = 53.2

    The value of سس is 53.2\boxed{53.2}.

Part 2

y=0.14195+0.00286xy = 0.14195 + 0.00286x

Regression Analysis

We are given the following data for Year (xx) and Exports (yy):

Year (xix_i)Exports (yiy_i)xi2x_i^2xiyix_i y_i
20.1540.30
70.16491.12
110.171211.87
130.181692.34
160.192563.04
Sum0.855998.67
  1. Calculate the sums.

    n=5n = 5
    ∑xi=2+7+11+13+16=49\sum x_i = 2 + 7 + 11 + 13 + 16 = 49
    ∑yi=0.15+0.16+0.17+0.18+0.19=0.85\sum y_i = 0.15 + 0.16 + 0.17 + 0.18 + 0.19 = 0.85
    ∑xi2=4+49+121+169+256=599\sum x_i^2 = 4 + 49 + 121 + 169 + 256 = 599
    ∑xiyi=0.30+1.12+1.87+2.34+3.04=8.67\sum x_i y_i = 0.30 + 1.12 + 1.87 + 2.34 + 3.04 = 8.67

  2. Calculate the slope (bb) of the regression line y=a+bxy = a + bx.

    The formula for bb is:

    b=n(∑xiyi)−(∑xi)(∑yi)n(∑xi2)−(∑xi)2b = \frac{n(\sum x_i y_i) - (\sum x_i)(\sum y_i)}{n(\sum x_i^2) - (\sum x_i)^2}

    Substitute the values:

    b=5(8.67)−(49)(0.85)5(599)−(49)2b = \frac{5(8.67) - (49)(0.85)}{5(599) - (49)^2} b=43.35−41.652995−2401b = \frac{43.35 - 41.65}{2995 - 2401} b=1.7594b = \frac{1.7}{594} b≈0.00286195b \approx 0.00286195
  3. Calculate the y-intercept (aa) of the regression line.

    The formula for aa is:

    a=∑yi−b(∑xi)na = \frac{\sum y_i - b(\sum x_i)}{n}

    Substitute the values:

    a=0.85−(0.00286195)(49)5a = \frac{0.85 - (0.00286195)(49)}{5} a=0.85−0.140235555a = \frac{0.85 - 0.14023555}{5} a=0.709764455a = \frac{0.70976445}{5} a≈0.14195289a \approx 0.14195289
  4. Write the regression equation.

    The regression equation is y=a+bxy = a + bx.

    y=0.14195+0.00286xy = 0.14195 + 0.00286x

    The regression equation is y=0.14195+0.00286x\boxed{y = 0.14195 + 0.00286x}.

Part 30.19919

Forecast exports for year 20

  1. Use the regression equation and substitute x=20x = 20.

    y20=0.14195289+0.00286195(20)y_{20} = 0.14195289 + 0.00286195(20) y20=0.14195289+0.057239y_{20} = 0.14195289 + 0.057239 y20=0.19919189y_{20} = 0.19919189
  2. Round the forecast to an appropriate number of decimal places.

    y20≈0.19919y_{20} \approx 0.19919

    The forecast for exports for year 20 is 0.19919\boxed{0.19919}.

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