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2 years ago

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The heights (in cm) and arm spans (in cm) of 31 students were measured. The association between x(height) and y(arm span) is sho

wn in the scatterplot below. The equation of the line of best fit modeling this relationship is given.
Y=4.5x + 0.977x

Mike is 175cm tall and George is 170cm tall.
Using the model, how many centimeters longer would we expect Mike's arm span to be than George's?

____Cm

1 answer:

abruzzese [7]2 years ago

8 0

Answer:

27.385 cm longer would we expect Mike's arm span to be than George's.

Step-by-step explanation:

We have to find how many centimeters longer would we expect Mike's arm span to be than George's using the equation:

y= 4.5x + 0.977 x

where y= arm span

and x= height

Given:

Mike's height=x = 175 cm

so Mike's arm span= y= 4.5 x + 0.977 x

= 4.5* (175) + 0.977* (175)

= 787.5 + 170.975

= 958.475 cm

George's height = x = 170 cm

so George's arm span= y= 4.5 x + 0.977 x

= 4.5* (170) + 0.977* (170)

= 765 + 166.09

= 931.09 cm

Mike's arm span longer than George's = Mike's arm span - George's arm span

= 958.475 - 931.09

= 27.385 cm.

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2 years ago

Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

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Answer:

Step-by-step explanation:

The equation of the sphere, centered a the origin is given by $x^2+y^2+z^2 = 64$ . Then, when z=4, we get

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c) We will use the previous analysis to define the limits in cartesian and polar coordinates. At first, we now that x varies from $-4\sqrt[]{3}$ up to $4\sqrt[]{3}$ . This is by taking y =0 and seeing the furthest points of x that lay on the circle. Then, we know that y varies from $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , this is again because y must lie in the interior of the circle we found. Finally, we know that z goes from 4 up to the sphere, that is , z goes from 4 up to $\sqrt[]{64-x^2-y^2}$

Then, the triple integral that gives us the volume of D in cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Recall that the cylindrical coordinates are given by $x=r\cos \theta, y = r\sin \theta,z = z$ , where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that $x^2+y^2 = r^2$ . WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ goes from 0 to $2\pi$ . Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

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2 years ago

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Step-by-step explanation:

The height of the ball at a time of 3.1 seconds is given by Charlize's equation as ...

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The well is about 151 feet deep.

_____

<em>Comment on this measurement</em>

The speed of sound is sufficiently slow that it will make a difference in the time Charlize records. She would get a more accurate measurement if she were able to <em>see</em> the ball hit bottom.

5 0

2 years ago