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3 years ago

Explain the difference between accuracy and precision

2 answers:

7 0

Precision is a description of random errors, a measure of statistical variability. Accuracy has 2 different definitions but the more common of the 2 is accuracy is systematic errors or errors that you control

nata0808 [166]3 years ago

3 0

Precision is more like consistency. Take a rifle range for example. Sure, all the shots may be off centered, but if they're tightly grouped together, it shows that the shooter had precision in their shooting. Accuracy is how close to spot on the attempts or shots generally are. They would end up more towards the center, but a little more spread out.

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(W+4)(3w-2) please solveeed

Shkiper50 [21]

Step-by-step explanation:

${:}\longrightarrow$ $\sf (w+4)(3w-2)$

${:}\longrightarrow$ $\sf w(3w-2)+4 (3w-2)$

${:}\longrightarrow$ $\sf 3w^2-2w+12w-8$

${:}\longrightarrow$ $\sf 3w^2+10w-8$

7 0

3 years ago

Read 2 more answers

After selecting 180 of its employees to from a new division, a company discovers that engineers outnumber analysts by a 3:1 rati

nikdorinn [45]

Answer:

I think its 90

Step-by-step explanation:

3x plus 1x is 4x divide 180 but 4 and get 45 so that meab there are 135 and 45

you need 90 more analysts to have a 1 to 1 ratio

3 0

2 years ago

Pls help

valina [46]

Answer:

Domain: {1, 2, 3, 4}

Step-by-step explanation:

The domain of the graph (input values) is the number of pizzas which are plotted on the x-axis while the range (output values) is the cost of pizza, plotted on the y-axis (vertical axis)

The domain therefore would consists of each x-coordinate that represent each point on the graph, which are {1, 2, 3, 4}

6 0

2 years ago

Nine runners are entered in a 100-meter dash for which a gold, silver, and bronze medal will be awarded for first, second, and t

DanielleElmas [232]

Answer:

504

Step-by-step explanation

9 people can get first x 8 runners left to win second x 7 who can still get third

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5C%5C%20%7B%20%5Crm%7BIf%20%20%5C%3A%20%7D%7B%5Ctt%7BA%20%3D%20%5Cleft%20%5B%20%5C%3A%20%5

RUDIKE [14]

Answer:

Given:

$\\ {\tt{A = \left [ \: \begin{array}{ c c c} \: \: \: \: 2 & -3 & - 5 \\ - 1 & \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: 1& - 3 & - 4\end{array} \: \: \: \right] \: and \: B = \left[\begin{array}{c c c} \: \: \: \: 2 & - 2 & - 4 \\ - 1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & - 2 & - 3\end{array} \right]}}$

Matrix A is of order 3 × 3 and Matrix B is of order 3 × 3

To Show:

Matrix AB = A , BA = B

Formula Used:

$\\ { \rm{ \left[ \begin{array}{c c c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array} \right] \times \left[ \begin{array}{c c c} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array} \right] = \: \left[ \begin{array}{c c c} \: \: a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} & a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32} & a_{11}b_{13} + a_{12}b_{23} + a_{13}b_{33} \\ \: \: a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} & a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32}&a_{21}b_{13} + a_{22}b_{23} + a_{23}b_{33} \\ \: \: a_{31}b_{11} + a_{32}b_{21} + a_{33}b_{31}&a_{31}b_{12} + a_{32}b_{22} + a_{33}b_{32} & a_{31}b_{13} + a_{32}b_{23} + a_{33}b_{33}\end{array} \right]}}$

If A is a matrix of order a×b and B is a matrix of order c×d , then matrix AB exits and is of order a×d , if and only if b = c.

$\: \:$

If A is a matrix of order a×b and B is a matrix of order c×d , then matrix BA exits and is of order c×b , if and only if d = a.

$\: \:$

For Matrix AB, a = 3, b = c = 3, d = 3 , thus Matrix AB is of order 3×3

$\\ { \rm{Matrix \: AB = \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: 1 & -3 & -4 \end{array} \right] \times \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right] = \left[\begin{array}{c c c} \: \: \: \: 2(2)-3(-1)-5(1) & \: \: \: \: 2(-2)-3(3)-5(-2) & \: \: \: \: 2(-4)-3(4)-5(-3) \\ -1(2)+4(-1)+5(1) & -1(-2)+4(3)+5(-2) & -1(-4)+4(4)+5(-3)\\ \: \: \: \: 1(2)-3(-1)-4(1) & \: \: \: \: 1(-2)-3(3)-4(-2) & \: \: \: \: 1(-4)-3(4)-4(-3) \end{array} \right]}}$

$\\ {\rm{Matrix \: AB = \left[ \begin{array}{c c c} \: \: \: \: 4 + 3 - 5 & \: \: \: -4 -9 + 10 & \: \: -8 - 12 + 15 \\ -2 - 4 + 5 & \: \: \: \: \: \: + 2 + 12 - 10 & \: \: \: \: \: 4 + 16 - 15 \\ \: \: \: \: \: 2 + 3 - 4 & \: \: -2 - 9 + 8 & -4 - 12 + 12 \end{array} \right] = \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: \: 4 & \: \: \: 5 \\ \: \: \: \: 1 & -3 & -4 \end{array} \right]}}$

$\\ {\rm{Matrix \: AB = \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: \: 4 & \: \: \: 5 \\ \: \: \: \: 1 & -3 & -4 \end{array} \right] = Matrix \: A}}$

<h3>Matrix AB = Matrix A</h3><h2>____________________________________________________________________</h2>

For Matrix BA, a = 3, b = c = 3, d = 3 , thus Matrix BA is of order 3 × 3

$\\ {\rm{Matrix \: BA = \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right] \times \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: \: 1 & -3 & -4 \end{array} \right]}}= \left[ \begin{array}{c c c} \: \: \: \: \: 2(2) -2(-1) -4(1) & \: \: \: \: 2(-3) -2(4) -4(-3) & \: \: \: \: 2(-5) -2(5) -4(-4) \\ \: -1(2) + 3(-1) + 4(1) & -1(-3) + 3(4) +4(-3) & -1(-5) + 3(5) + 4(-4) \\ \: \: \: \: \: 1(2) -2(-1) -3(1) & \: \: \: \: 1(-3) -2(4) -3(-3) & \: \: \: \: 1(-5) -2(5) -3(-4) \end{array} \right]$

$\\{ \rm{Matrix \: BA = \left[ \begin{array}{c c c} \: \: \: \: 4 + 2 - 4 & \: \: \: -6 -8 + 12 & \: \: -10 - 10 + 16 \\ -2 - 3 + 4 & \: \: \: \: \: \: +3 + 12 - 12 & \: \: \: \: \: + 5 + 15 - 16 \\ \: \: \: \: \: 2 + 2 - 3 & \: -3 -8 +9 & \: \: \: \: \: -5 -10 + 12 \end{array} \right] = \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right]}}$

$\\{ \rm{Matrix \: BA = \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right] = Matrix \: B}}$

<h3>Matrix BA = Matrix B</h3><h2>____________________________________________________________________</h2>

3 0

2 years ago