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Vaselesa [24]
3 years ago
13

Karen is studying the relationship between the time spent exercising per day and the time spent outside per day and has collecte

d the data shown in the table. The line of best fit for the data is y^=0.16x+45.5. Assume the line of best fit is significant and there is a strong linear relationship between the variables. Exercising (Minutes) 7090110130 Time Spent Outside (Minutes) 57606267 According to the line of best fit, what would be the predicted number of minutes spent outside for someone who spent 68 minutes exercising? Round your answer to two decimal places, as needed.
Mathematics
1 answer:
wariber [46]3 years ago
7 0

Answer:

56.38=

Step-by-step explanation:

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A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500
Vedmedyk [2.9K]

Answer:

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

Given:

Rate at which the woman is walking,\frac{d(w)}{dt} = 3 ft/sec

Rate at which the man is walking,\frac{d(m)}{dt} = 2 ft/sec

Collective rate of both, \frac{d(m+w)}{dt} = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min  = 20 min

Now,

Distance traveled by man and woman are m and w ft respectively.

⇒ m=2\ ft/sec=2\times \frac{60}{min} \times 20\ min =2400\ ft

⇒ w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\  ft

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate \frac{dh}{dt} .

Lets calculate h.

Applying Pythagoras formula.

⇒ h^2=(m+w)^2+500^2  

⇒ h=\sqrt{(2400+2700)^2+500^2} = 5124.45

Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒ h^2=(m+w)^2+500^2

⇒ 2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt}  + \frac{d(500)}{dt}

⇒ \frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}                         ...as \frac{d(500)}{dt}= 0

⇒ Plugging the values.

⇒ \frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}                       ...as \frac{d(m+w)}{dt} = 5 ft/sec

⇒ \frac{d(h)}{dt} =4.9761  ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

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3 years ago
The amount of space in the dog carrier shown can be found by multiplying it's width length and height write the amount of space
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V = 8 ft³
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