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DochEvi [55]
3 years ago
6

A) choose the linear model that passes through the most data points on the scatterplot.

Mathematics
1 answer:
irina1246 [14]3 years ago
8 0
Answer: b

Explanation:
The line of best fit to a scattergram is obtained in linear regression analysis by minimizing the sum of the squared errors.
For example, in the diagram shown below, there are n data points in the scattergram.
The error for the i-th data point is e_{i}.
The coefficients (a and b) for the line of best fit are determined using calculus, to minimize \sum_{i=1}^{n} e_{i}^{2}.

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Is seven times the difference of a number and 4 no more than 10?

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Find the slope of the line through the two points below.<br> (-5, 3) and (-1,0)
Maru [420]

Answer:

m = -3/4

Step-by-step explanation:

Hope this helps:)

3 0
3 years ago
Help me out with these 2 questions for 15 points.
jok3333 [9.3K]

Step-by-step explanation:

The time dilation formula is given by

F(t) = \dfrac{t}{\sqrt{1-v^2}}

where t is the time measured by the moving observer and F(t) is the time measured by the stationary earth-bound observer and <em>v</em><em> </em> is the velocity of the moving observer expressed as a fraction of the speed of light.

a) If the observer is moving at 80% of the speed of light and observes an event that lasts for 1 second, a stationary observer will see the same event occurring over a time period of

F(t) = \dfrac{1\:\text{s}}{\sqrt{1-(0.8)^2}}

\:\:\:\:\:\:\:=\dfrac{1\:\text{s}}{\sqrt{1-(0.8)^2}} =\dfrac{1\:\text{s}}{\sqrt{1-(0.64)}}

\:\:\:\:\:\:\:=1.67\:\text{s}

This means that any event observed by this moving observer will be seen by a stationary observer to occur 67% longer.

b) Given:

t = 1 second

F(t) = 2 seconds

We need to find the speed of the observer such that an event seen by this observer will occur twice as long as seen by a stationary observer. Move the term containing the radical to the left side so the equation becomes

\sqrt{1-v^2} = \dfrac{t}{F(t)}

Take the square of both sides, we get

1 - v^2 =\dfrac{t^2}{F^2(t)}

Solving for <em>v</em>, we get

v^2 = 1 - \dfrac{t^2}{F^2(t)}

or

v = \sqrt{1 - \dfrac{t^2}{F^2(t)}}

Putting in the values for t and F(t) we get

v = \sqrt{1 - \dfrac{(1\:\text{s})^2}{(2\:\text{s})^2}}

v = \sqrt{1 - \dfrac{1}{4}} = \sqrt{0.75}

\:\:\:\:=0.866

This means that the observer must moves at 86.6% of the speed of light.

6 0
3 years ago
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