45 is 35% of what number?
2 answers:
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Answer:
-95.78
Step-by-step explanation:
As the researcher decided to make the number of parties attended per week the explanatory variable, this would be variable x in the regression line, and of course, the variable y would be the number of text messages sent per day.
After constructing the linear regression equation, the researcher found that an approximate value for the actual value of y could be represented by the line
Since this is an approximate value, it is not expected that it coincides with the actual value of y. We define then the residual for each value of x as the difference between the actual value of y and the approximation for the given x.
For the value x = 2 (the student attended 2 parties that week) the actual value of y is 20 (the student sent 20 text messages per day that week).
The approximate value of y would be according to the regression line
Hence, the residual value for x=2 would be
9514 1404 393
Answer:
m ≥ -√5
Step-by-step explanation:
If g(x) = f(|x|) for x∈i, then g(x) = f(x) for x∈ℝ: x ≥ 0.
f'(x) is 5th-degree, so f(x) is 6th-degree, meaning it is generally U-shaped. Since we're only concerned with x ≥ 0, we want to make sure f'(x) has no real zeros of odd multiplicity such that x > 0. The given factors of f'(x) make it have real zeros at x = -3 and x = -1.
For the last factor, (x² +2mx +5) to have no positive real zeros of odd multiplicity, we must have m ≥ 0 or the discriminant ≤ 0. The discriminant is ...
d = b² -4ac = (2m)² -4(1)(5) = 4m² -20 . . . . . discriminant of the last factor
d ≤ 0 . . . . . . . . . . the condition for no real zeros
4m² -20 ≤ 0
m² -5 ≤ 0 . . . . . . divide by 4
m² ≤ 5 . . . . . . . . .add 5
|m| ≤ √5 . . . . . . . take the square root
This tells us there will be a positive real zero of multiplicity 2 in f'(x) when m = -√5, and there will be no positive real zeros for -√5 < m < 0
There will be no odd-multiplicity positive real zeros in the derivative function f'(x) as long as m ≥ -√5. This means the slope of f(x) is non-negative for x ≥ 0, hence f(|x|) has its only minimum at x=0.
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<em>Additional comment</em>
The multiplicity of the zeros of f'(x) is important because the derivative will only change sign where the multiplicity is odd. When the discriminant of (x²+2mx+5) is zero, the associated positive real zero will have multiplicity 2, hence f'(x) will not change sign there.
