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alexandr1967 [171]
3 years ago
15

With Obesity on the rise, a Doctor wants to see if there is a linear relationship between the Age and Weight and estimating a pe

rson's Systolic Blood Pressure. Using the Estimated Regression equation, Estimate Systolic BP when some if 39 years old and they weigh 143 pound.
A. 114.785.
B. 115.532.
C. 122.471.
D. 112.569
Systolic BP Age in Yrs. Weight in lbs.
132 52
143 59 184
153 67 194
162 73 168
154 64 196
168 74 220
137 54 188
149 61 188
159 65 207
128 46 167
166 72 217
135 52 187
148 61 161
148 61 189
161 65 205
126 45 161
167 75 215

Mathematics
1 answer:
Romashka [77]3 years ago
8 0

Answer:

112.569 ( D )

Step-by-step explanation:

Applying the estimated Regression Equation

y = b1X1 + b2X2 + a

b1 = ((SPX1Y)*(SSX2)-(SPX1X2)*(SPX2Y)) / ((SSX1)*(SSX2)-(SPX1X2)*(SPX1X2)) = 596494.5/635355.88 = 0.93884

b2 = ((SPX2Y)*(SSX1)-(SPX1X2)*(SPX1Y)) / ((SSX1)*(SSX2)-(SPX1X2)*(SPX1X2)) = 196481.5/635355.88 = 0.30925

a = MY - b1MX1 - b2MX2 = 149.25 - (0.94*61.31) - (0.31*193.88) = 31.73252

y = 0.939X1 + 0.309X2 + 31.733

For x1 ( age ) =39, and  x2(weight) =143

y = (0.93884*39) + (0.30925*143) + 31.73252= 112.569

where

Sum of X1 = 981

Sum of X2 = 3102

Sum of Y = 2388

Mean X1 = 61.3125

Mean X2 = 193.875

Mean Y = 149.25

attached is the Tabular calculation of the required values needed for estimated regression equation

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A rock thrown vertically upward from the surface of the moon at a velocity of 36​m/sec reaches a height of s = 36t - 0.8 t^2 met
Verdich [7]

Answer:

a. The rock's velocity is v(t)=36-1.6t \:{(m/s)}  and the acceleration is a(t)=-1.6  \:{(m/s^2)}

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

  • Velocity is defined as the rate of change of position or the rate of displacement. v(t)=\frac{ds}{dt}
  • Acceleration is defined as the rate of change of velocity. a(t)=\frac{dv}{dt}

a.

The rock's velocity is the derivative of the height function s(t) = 36t - 0.8 t^2

v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t

The rock's acceleration is the derivative of the velocity function v(t)=36-1.6t

a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6

b. The rock will reach its highest point when the velocity becomes zero.

v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = \frac{405}{2} =202.5 \:m

To find the time it reach half its maximum height, we need to solve

36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45

The rock is aloft for 45 seconds.

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Coordinates of B → (12, 12)

If a point (x, y) is dilated by a scale factor 'k' about the origin, rule to be followed,

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(x, y) → (\frac{2}{3}x, \frac{2}{3}y)

By this rule coordinates of the image points of A and B will be,

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B(12, 12) → B'(\frac{12\times 2}{3}, \frac{12\times 2}{3})

             → B'(8, 8)

Now we can get the image of segment AB after dilation by a scale factor of \frac{2}{3}.

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