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Ainat [17]
3 years ago
13

The graph below plots the values of y for different values of x: plot the ordered pairs 1, 3 and 2, 4 and 3, 9 and 4, 7 and 5, 2

and 6, 18 What does a correlation coefficient of 0.25 say about this graph? x and y have a strong, positive correlation x and y have a weak, positive correlation x and y have a strong, negative correlation x and y have a weak, negative correlation
Mathematics
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

x and y have a weak, positive correlation

Step-by-step explanation:

:)

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Please help me! I need it
Studentka2010 [4]

Answer:

About 0.5

Step-by-step explanation:

By the law of sines:

\dfrac{2}{\sin 105}=\dfrac{c}{\sin 15} \\\\\\c=\dfrac{2}{\sin 105}\cdot \sin 15\approx 0.5cm

Hope this helps!

3 0
3 years ago
What the median of 17, 24, 8, 19, 6, 34, 10, 28, 48, 12<br>​
BartSMP [9]

The median is 48 because it is the biggest number and the biggest number is the median
4 0
3 years ago
The body of a 154-pound person contains approximately 2×10^2 milligrams of gold and 6×10^3 milligrams of aluminum. Based on this
hram777 [196]

Answer:

2x10^2 and 6x 10^

Step-by-step explanation:

Hope I helped UwU

3 0
3 years ago
ATB is a tangent to the circle centre O. Find the values of x y and z in each case.
Triss [41]

Answer:

Step-by-step explanation:

divde then add 67b

then complete

7 0
3 years ago
Find the values of λ for which the determinant is zero.
elena55 [62]

Answer:

Determinant are special number that can only be defined for square matrices.

Step-by-step explanation:

Determinant are particularly important for analysis. The inverse of a matrix exist, if the determinant is not equal to zero.

How to find determinant

For a 2×2 matrix

det ( \left[\begin{array}{cc}x&y\\a&z\end{array}\right] ) = xz-ay

For a 3×3 matrix

we first decompose it to 2×2

det (\left[\begin{array}{ccc}k&l&m\\o&p&q\\r&s&t\end{array}\right] )\\\\= k*det(\left[\begin{array}{cc}p&q\\s&t\end{array}\right] ) - l*det(\left[\begin{array}{cc}o&q\\r&t\end{array}\right] ) + m*det(\left[\begin{array}{cc}o&p\\r&s\end{array}\right] ) \\\\=k(pt-sq) - l(ot-rq) + m(os-rp)

Example

Find the values of λ for which the determinant is zero

\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right]

det(\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right])\\\\= s*det(\left[\begin{array}{cc}s&-1\\-1&1\end{array}\right] ) - (-1)*det(\left[\begin{array}{cc}-1&-1\\0&1\end{array}\right] ) + 0*det(\left[\begin{array}{cc}-1&s\\0&-1\end{array}\right] )\\\\= s(s(1)-(-1*-1)) - (-1)(-1*1 - (-1*0)) + 0\\= s(s - 1)) + 1(-1 + 0) \\=s^{2} -s-1

Equating the determinant to zero

s^{2} -s-1 =0\\

s = \frac{1}{2} * (1 ±5 )

s = 1.61 or -0.61

5 0
3 years ago
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