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Nastasia [14]
3 years ago
9

PLEASE HELP ME ASAP (ANSWER BOTH) THANK YOU!

Mathematics
1 answer:
Svetach [21]3 years ago
8 0

Answer:

1/3 is the slope but it does not show the graph so I can't give you the answer

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3 years ago
The class sizes of the introductory psychology courses at a college are shown below 121,134,106,93,149,130,119,128 the college a
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Center : Mean  Before the introduction of the new course, center = average(121,134,106,93,149,130,119,128) = 122.5  After the introduction of the new course, center =  average(121,134,106,93,149,130,119,128,45) = 113.9  The center has moved to the left (if plotted in a graph) because of the low intake for the new course.  Spread before introduction of the new course :  Arrange the numbers in ascending order:  (93, 106,119, 121), (128, 130,134, 149) Q1=median(93,106,119,121) = 112.5 Q3=median(128,130,134,149) = 132  Spread = Interquartile range = Q3-Q1 = 19.5  After addition of the new course,

 (45,93, 106,119,) 121, (128, 130,134, 149)
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3 years ago
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Determine whether the points lie on straight line.
wel

Answer:

A) Non-collinear- does not lies on straight line

B) Collinear- lie on straight line

Step-by-step explanation:

We have to check collinearity of three points.

The points (x_1, y_1, z_1),(x_2, y_2, z_2),(x_3, y_3, z_3) are said to be collinear if,

\left[\begin{array}{ccc}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{array}\right] = 0

A) A(2, 5, 3), B(3, 7, 1), C(1, 4, 4)

\left[\begin{array}{ccc}2&5&3\\3&7&1\\1&4&4\end{array}\right] \\\\=2(28-4)-5(12-1)+3(12-7)\\= 8 \neq 0

Thus, the given points are not collinear.

B) D(0,-5,5), E(1,-2,4), F(3,4,2)

\left[\begin{array}{ccc}0&-5&5\\1&-2&4\\3&4&2\end{array}\right] \\\\=0(-4-16)+5(2-12)+5(4+6)\\=0

Thus, the given points are collinear.

6 0
3 years ago
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustra
Rainbow [258]

Answer:

x = t

y = 1 - t

z = 2t

Step-by-step explanation:

Given

x=t

y=e^{-t}

z=2t-t^2

(0, 1, 0)

The vector equation is given as:

r(t) = (x,y,z)

Substitute values for x, y and z

r(t) = (t,\ e^{-t},\ 2t - t^2)

Differentiate:

r'(t) = (1,\ -e^{-t},\ 2 - 2t)

The parametric value that corresponds to (0, 1, 0) is:

t = 0

Substitute 0 for t in r'(t)

r'(t) = (1,\ -e^{-t},\ 2 - 2t)

r'(0) = (1,\ -e^{-0},\ 2 - 2*0)

r'(0) = (1,\ -1,\ 2 - 0)

r'(0) = (1,\ -1,\ 2)

The tangent line passes through (0, 1, 0) and the tangent line is parallel to r'(0)

It should be noted that:

The equation of a line through position vector a and parallel to vector v is given as:

r(t) = a + tv

Such that:

a = (0,1,0) and v = r'(0) = (1,-1,2)

The equation becomes:

r(t) = (0,1,0) + t(1,-1,2)

r(t) = (0,1,0) + (t,-t,2t)

r(t) = (0+t,1-t,0+2t)

r(t) = (t,1-t,2t)

By comparison:

r(t) = (x,y,z) and r(t) = (t,1-t,2t)

The parametric equations for the tangent line are:

x = t

y = 1 - t

z = 2t

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