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4 years ago

Which of these is a true statement

Mathematics

2 answers:

Vikki [24]4 years ago

7 0

Answer:

d may cause f (option D)

Step-by-step explanation:

We can define a correlation as a type of association between two variables. There can be a positive, a negative or no correlation between two variables.

A correlation is a condition for causation, but correlation does not imply any definite causation.

Hence, the answer here is option D : d may cause f.

KengaRu [80]4 years ago

5 0

The true statement is d, d may cause f.

Explanation:
Strong correlation doesn't always mean causation. For example, there is correlation between time and distance. Time doesn't cause the distance. If time added, the distance is much longer. From the example above, correlation doesn't always mean causation. But for any other example, like, heat has strong correlation with ice melts. The heat is the cause for ice to melt, This correlation might be causation.

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What is the answer of this?

amid [387]

So the modal value is most commonly known as mode, or the most common number in a data set. So the most common number in this stem-and-leaf plot is 137

The modal value is 137
Hopes this helps!

5 0

4 years ago

Charlie is at a small airfield watching for the approach of a small plane with engine trouble. He sees the plane at an angle of

Lera25 [3.4K]

Check the picture.

Let the given points represent the following:

F : Charlie's feet
E: Charlie's eyes
P: location of the plane
G: the foot of the perpendicular drawn from P to the ground
A: a point on PG, such that |GA|=|FE|=5.2 feet.

Clearly EAP is a right triangle, with m(AEP)=32°, side PA=1,700-5.2=1694.8 feet.

<span>the ground distance from Charlie to the plane is |FG|=|EA|

from right angle trigonometry, we know that:

</span> $tan32= \frac{PA}{AE}$
<span>
</span> $0.625= \frac{1694.8}{AE}$
<span>
</span> $AE= \frac{1694.8}{0.625}=2711.68$ feet
<span>

Answer: 2711.68 feet</span>

3 0

3 years ago

Given that angle a = 54° and angle b = 48°, work out 3.

natima [27]

9514 1404 393

Answer:

x = 78°

Step-by-step explanation:

The sum of the interior angles of a triangle is 180°.

x + a + b = 180°

x = 180° -a -b . . . . . . . . add -a-b to both sides

x = 180° -54° -48°

x = 78°

7 0

3 years ago

Find the slope of the line that passes through the points (2, 4) and (6, 12).

Gnesinka [82]

Answer:

slope of the line passing through these coordinates is 2

Step-by-step explanation:

slope of line :-

=》

$\frac{y2 - y1}{x2 - x1}$

=》

$\frac{12 - 4}{6 - 2}$

=》

$\frac{8}{4}$

=》

$2$

so, the slope of the line is 2

6 0

3 years ago

3. for each item, decide whether or not the given expression is defined. for each item that is defined, compute the result. (a)

Sati [7]

The results of given matrices can be obtained using matrix multiplication.

<h3>Find the results of the given matrices:</h3>

Here in the question it is given that,

A = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$ , B = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ , C = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ , D = $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$ ,

E = $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$ , F = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

We have to find AB, BC, CA, CD, $C^{T} A^{T}$ , F², $BD^{T}$ , $A^{T} A$ and FE.

AB = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$

a₁₁ = 1×2 + (-1)×5 + 2×4 = 5, a₁₂ = 1×(-1) + (-1)×1 + 2×6 = 10, a₁₃ = 1×3 + (-1)×2 + 2×(-2) = -3, a₂₁ = 3×2 + 1×5 + 4×4 = 27, a₂₂ = 3×(-1) + 1×1 + 4×6 = 22, a₂₃ = 3×3 + 1×2 + 4×(-2) = 3

AB = $\left[\begin{array}{ccc}5&10&-3\\27&22&3\end{array}\right]$

BC = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$

a₁₁ = 2×1 + (-1)×(-1) + 3×2 = 9, a₂₁ = 5×1 + 1×(-1) + 2×2 = 8, a₃₁ = 4×1 + 6×(-1) + (-2)×2 = -6

BC = $\left[\begin{array}{ccc}9\\8\\-6\end{array}\right]$

CA, CA is not defined since dimension of the matrices are 3×1 and 2×3

$A^{T}E = \left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right]\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$

a₁₁ = 1×(2-i) + 3×(-i) = 2-4i, a₁₂ = 1x(1+i) + 3×(2+4i) = 7+13i, a₂₁ = -1×(2-i) + 1×(-i) = -2, a₂₂ = -1×(1+i) + 1×(2+4i) = 1+3i, a₃₁ = 2×(2-i) + 4×(-i) = 4-6i, a₃₂ = 2×(1+i) + 4×(2+4i) = 10+18i

$A^{T}E = \left[\begin{array}{ccc}2-4i&7+13i\\-2&1+3i\\4-6i&10+18i\end{array}\right]$

CD = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$

a₁₁ = 1×2 = 2, a₁₂ = 1×(-2) = -2, a₁₃ = 1×3 = 3, a₂₁ = -1×2 = -2, a₂₂ = -1×(-2) = 2, a₂₃ = -1×3 = -3,a₃₁= 2×2 = 4, a₃₂ = 2×(-2) = -4, a₃₃ = 2×3 = 6

CD = $\left[\begin{array}{ccc}2&-2&3\\-2&2&-3\\4&-4&6\end{array}\right]$

$C^{T} A^{T} =\left[\begin{array}{ccc}1&-1&2\end{array}\right]\left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right]$

a₁₁ = 1×1 + (-1)×(-1) + 2×2 = 6, a₁₂ = 1×3 + (-1)×1 + 2×4 = 10

$C^{T}A^{T}=\left[\begin{array}{ccc}6&10\end{array}\right]$

F² = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

a₁₁ = i×i + (1-3i)×0 = -1,a₁₂ = i×(1-3i) + (1-3i)×(4+i) = 10-10i, a₂₁= 0×i + (4+i)×0 = 0, a₂₂ = 0×(1-3i) + (4+i)×(4+i) = 15+8i

F² = $\left[\begin{array}{ccc}-1&10-10i\\0&15+8i\end{array}\right]$

$BD^{T}=\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]\left[\begin{array}{ccc}2\\-2\\3\end{array}\right]$

a₁₁ = 2×2 + (-1)×(-2) + 3×3 = 15, a₂₁ = 5×2 + 1×(-2) + 2×3 = 14, a₃₁ = 4×2 + 6×(-2) + (-2)×3 = -10

$BD^{T}= \left[\begin{array}{ccc}15\\14\\-10\end{array}\right]$

$A^{T} A=\left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right] \left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$

a₁₁ = 1×1 + 3×3 = 10, a₁₂ = 1×(-1) + 3×1 = 2, a₁₃ = 1×2 + 3×4 = 14, a₂₁ = -1×1 + 1×3 = 2, a₂₂ = -1×(-1) + 1×1 = 2, a₂₃ = -1×2 + 1×4 = 2, a₃₁ = 2×1 + 4×3 = 14, a₃₂ = 2×(-1) + 4×1 = 2, a₃₃ = 2×2 + 4×4 = 20

$A^{T} A=\left[\begin{array}{ccc}10&2&14\\2&2&2\\14&2&20\end{array}\right]$

FE = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$ $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$

a₁₁ = i×(2-i) + (1-3i)×(-i) = -2+i, a₁₂ = i×(1+i) + (1-3i)×(2+4i) = 13-i, a₂₁ = 0×(2-i) + (4+i)×(-i) = 1-4i, a₂₂ = 0×(1+i) + (4+i)×(2+4i) = 4+18i

FE = $\left[\begin{array}{ccc}-2+i&13-i\\1-4i&4+18i\end{array}\right]$

Hence we can obtain the results of the required matrices using matrix multiplication.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Let A = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$ , B = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ , C = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ , D = $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$ , E = $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$ , F = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

For each item, decide whether or not the given expression is defined. for each item that is defined, compute the result.

AB, BC, CA, CD, $C^{T} A^{T}$ , F², $BD^{T}$ , $A^{T} A$ and FE

Learn more about matrix here:

brainly.com/question/28180105

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8 0

2 years ago