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2 years ago

Mean median mode range of 13,11,8,8,12,8

Mathematics

2 answers:

igomit [66]2 years ago

7 0

I think the median is 8 or 11

myrzilka [38]2 years ago

3 0

Mode (most common number) = 8
Median (# in middle when put all #s in order):
8 8 8 11 12 13 -> 8 and 11 are in the middle so find average of them = 9.5
Mean = 10

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Danna completes 2 math problems per minute. What is the start value

Yanka [14]

Answer:

Step-by-step explanation:1math problem in 30seconds

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2 years ago

Nick wanted to determine the length of one blade of a windmill. He stood at a point on the ground 440 feet from the windmill's b

mafiozo [28]

Answer:There are two right triangles. You need to find the opposite side (the height) of each of them. The difference between the two is the height of the blade, x.

You know the adjacent, 440, and you are looking for the opposite, so you need to use tan for both.

tan (30) = y / 440

y = 440 * tan (30) = 254.034...

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Step-by-step explanation:

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1 year ago

- 3x + 4 and passes through the point (2,5)

Anuta_ua [19.1K]

Answer:

y=(x+13)/3

Step-by-step explanation:

the question is not quite clear, i think u mean y= -3x + 4

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2 years ago

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

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No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$

Hence, the correct answer is option A) 1260

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3 years ago

All vectors are in Rn. Check the true statements below:

Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

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D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .

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3 years ago