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Brilliant_brown [7]
3 years ago
14

Question 2(Multiple Choice Worth 5 points) (08.02 MC)What is the mean absolute deviation for 2, 9, 1, 7, 8, and 9? 1 3 6 8

Mathematics
1 answer:
amm18123 years ago
3 0

Answer:

Mean Absolute Deviation = 3.

Step-by-step explanation:

The mean = (2 + 9 + 1 + 7 + 8 + 9) / 6

= 36/6

= 6.

Subtract  6 from each  number:

2 - 6 = -4 Absolute value = 4

9 - 6 = 3

1 - 6 = -5 Absolute value = 5

7 - 6 = 1

8 - 6 = 2

9 - 6 = 3

Total = 4 + 3 + 5 + 1 + 2 + 3 = 18

Mean Absolute Deviation = 18 / 6 = 3.

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saul85 [17]

The probability that if 5 tires are inspected, exactly 2 have a defect is 0.021

<h3>How to calculate the probability</h3>

The given parameters are:

p = 5\% --- the proportion that contains defect

n =5 --- the sample size

r =2 -- the selected sample

The probability is then calculated as:

P(x =r) = ^nC_r * p^r * (1 - p)^{n-r}

So, we have:

P(x =2) = ^5C_2 * (5\%)^2 * (1 - 5\%)^{5-2}

P(x =2) = 10 * (5\%)^2 * (1 - 5\%)^3

P(x =2) = 0.021

Hence, the probability that if 5 tires are inspected, exactly 2 have a defect is 0.021

Read more about probabilities at:

brainly.com/question/15246027

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2 years ago
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Answer:

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Step-by-step explanation:

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5 0
3 years ago
L :V --&gt; W is a linear transformation. Prove each of the following (a) ker L is a subspace of V. (b) range L is a subspace of
iragen [17]

Answer:

a) Assume that x,y\in\ker L, and \alpha is a scalar (a real or complex number).

<em>First. </em>Let us prove that \ker L is not empty. This is easy because L(0_V)=0_W, by linearity. Here, 0_V stands for the zero vector of V, and 0_W stands for the zero vector of W.

<em>Second.</em> Let us prove that \alpha x\in\ker L. By linearity

L(\alpha x) = \alpha L(x)=\alpha 0_W=0_W.

Then, \alpha x\in\ker L.

<em>Third. </em> Let us prove that y+ x\in\ker L. Again, by linearity

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b) Assume that u and v are in range of L. Then, there exist x,y\in V such that L(x)=u and L(y)=v.

<em>First.</em> Let us prove that range of L is not empty. This is easy because L(0_V)=0_W, by linearity.

<em>Second.</em> Let us prove that \alpha u is on the range of L.

\alpha u = \alpha L(x) = L(\alpha x) = L(z).

Then, there exist an element z\in V such that L(z)=\alpha u. Thus \alpha u is in the range of L.

<em>Third.</em> Let us prove that u+v is in the range of L.

u+v = L(x)+L(y) = L(x+y)=L(z).

Then, there exist an element z\in V such that L(z)= u +v. Thus u +v is in the range of L.

Notice that in this second part of the problem we used the linearity in the reverse order, compared with the first part of the exercise.

Step-by-step explanation:

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The answer to this question is about 12
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