We square the residuals when using the least-squares line method to find the line of best fit because we believe that huge negative residuals (i.e., points well below the line) are just as harmful as large positive residuals (i.e., points that are high above the line).
<h3>What do you mean by Residuals?</h3>
We treat both positive and negative disparities equally by squaring the residual values. We cannot discover a single straight line that concurrently minimizes all residuals. The average (squared) residual value is instead minimized.
We might also take the absolute values of the residuals rather than squaring them. Positive disparities are viewed as just as harmful as negative ones under both strategies.
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Let X be a random sample representing the score of a student in the American Chemical Society Examination.
P(450 < x < 500) = P((450 - 500)/(90/sqrt(25)) < z < (500 - 500)/(90/sqrt(25))) = P(-2.778 < z < 0) = P(z < 0) - P(z < -2.778) = P(z < 0) - [1 - P(z < 2.778)] = P(z < 0) + P(z < 2.778) - 1 = 0.5 + 0.99727 - 1 = 0.49727 = 49.7%
Answer:
3. BDE is congruent to BAC; corresponding angles postulate
4. B is congruent to B; reflexive property of equality
Step-by-step explanation:
I took the test.
Answer:
The graph of a linear equation is a straight line. The "solution" to a system of two linear equations is the point where the two lines cross. If the two lines are parallel, they never cross; hence parallel lines have no solution. Two lines are parallel if they have the same slope (the m value in y = mx+b). One of your equations is y = -2x + (you left the y-intercept out). The slope is -2. So any line with a slope of m = -2 will be parallel to this line and will not cross it. The second line also needs a different value of b, the y-intercept. Otherwise it is the same line and every point is a solution. So if your equation is:
y = -2x + 1
Then any equation of the form y = -2x + b, b≠1 will create a system with no solution. Hence the values of m and b are m = -2, b ≠ 1.
Answer:
-40 i think
Step-by-step explanation:
