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Irina18 [472]
3 years ago
7

When a trend line is drawn on a scatterplot, there are 4 points above the trend line. About how many points should be below the

trend line?
Mathematics
2 answers:
motikmotik3 years ago
6 0

Answer:

The number of points that should lie below the trend line is:

4.

Step-by-step explanation:

" The general rule of thumb is to have your trend line as close as possible to all data points, and with equal number of points above and below it "

The trend line is also called line of best fit.

It is given that:

When a trend line is drawn on a scatter plot, there are 4 points above the trend line.

Hence, the number of points that should lie below the trend line are:

4.

scoundrel [369]3 years ago
5 0
There is no exact rule for lines of best fit. However, in general, there should be roughly the same amount above as below. So, if we are following this rule, there should be about 4 below as well. 
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We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance
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Answer:

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

h(t) = -16\cdot t^{2}+256\cdot t+1680 (1)

Where:

h - Height above the ground, measured in feet.

t - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680

h(3\,s) = 2304\,ft

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at t = 5\,s and t = 7\,s:

h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680

h(5\,s) = 2560\,s

h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680

h(7\,s) = 2688\,s

We notice that the small object ascends in the given interval.

\Delta h = h(7\,s)-h(5\,s)

\Delta h = 128\,ft

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that h = 2640\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t-960=0 (2)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 10\,s and t_{2}= 6\,s

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that h = 0\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t +1680 = 0 (3)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 21\,s and t_{2} = -5\,s

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

8 0
3 years ago
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