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d1i1m1o1n [39]
3 years ago
6

The scatter plot shows the number of customers in a restaurant for hours of the dinner services on two different Saturday nights

. The line shown models this relationship, and x=0 represent 7:00pm
A. What does the value y-intercept represent in the context of the problem (shown inherent attachment)


B. What is the slope of the line?


C. Write an equation to represent the line of nest fit drawn


D. Wat is the number of customers the restaurant should expect at 8:30pm?

Mathematics
1 answer:
charle [14.2K]3 years ago
5 0

A. The value y-intercept represent in the context of the problem is 50.

B. The slope of the line is -10.

C. The equation to represent the line is y= -10x+50.

D. The number of customers in the restaurant at 8:30pm is 35.

<u>Step-by-step explanation:</u>

The equation of the line is given by y=mx+c.

y is the number of customers and x is the hours.

7.00 p.m. is considered as 0.

To find the slope m =\frac{rise}{run}.

rise = y_{2}-y_{1}.

run = x_{2}-x_{1}.

From the graph, choose two points. Let us take (0,50) and (4,10).

m=\frac{-40}{4}.

m=-10.

∴The equation can be written as y= -10x+c.

To find the y-intercept value c, substitute any point.

Let us substitute(0,50) for instance,

50=0+c.

c=50.

The equation for the given graph is y= -10x+50.

To find the customers at the restaurant at 8.30 pm.

we know that x=0 at 7.00 pm,

∴ x=1.5 can be considered as 8.30 pm.

Substitute x=1.5 in the line equation.

y= -10(1.5)+50.

y= -15-50.

y= 35.

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(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

(b) The velocity after 3 seconds is

v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}

(c) The particle is at rest when its velocity is zero:

\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

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\displaystyle \int_0^8 |v(t)|\,\mathrm dt

By definition of absolute value, we have

|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}

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