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2 years ago

Question 8 of 10 What is the median of this data set? 14, 18, 31, 34, 44, 50 A. 32.5 B. 33 C. 31.5 O O D. 31 SUBM

Mathematics

2 answers:

kozerog [31]2 years ago

8 0

Answer:

32.5

Step-by-step explanation:

median is the middle number, in this case the sum of the two middle number divided by 2

31 + 34 = 65/2

32.5

andreyandreev [35.5K]2 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely

navik [9.2K]

Answer:

Never

Step-by-step explanation:

The equations given are

2x1−6x2−4x3 = 6 ....... (1)

−x1+ax2+4x3 = −1 ........(2)

2x1−5x2−2x3 = 9 ..........(3)

the values of a for which the system of linear equations has no solutions

Let first add equation 1 and 2. Also equation 2 and 3. This will result to

X1 + (a X2 - 6X2) - 0 = 5

And

X1 + (aX2-5X2) + 2X3 = 8

Since X2 and X3 can't be cancelled out, we conclude that the value of a is never.

a unique solution,

Let first add equation 1 and 2. Also equation 2 and 3. This will result to

X1 + (a X2 - 6X2) - 0 = 5

And

X1 + (aX2-5X2) + 2X3 = 8

The value of a = never

infinitely many solutions.

Divide equation 1 by 2 we will get

X1 - 3X2 - 2X3 =3

Add the above equation with equation 3. This will result to

3X1 - 8X2 - 4X3 = 12

Everything ought to be the same. Since they're not.

Value of a = never.

4 0

3 years ago

HELP PLEASE !!!!!!!!!!

Arada [10]

Answer:

Answer B $(-\infty,1)U(1,2]$

Step-by-step explanation:

Notice that the quotient of f(x)/g(x) is:

$h(x)=\frac{\sqrt{2-x} }{x-1}$

therefore, this new function imposes conditions due to the fact that it has a square root in the numerator and a binomial in the denominator both with the variable x. Then, in order for the root in the numerator to be defined, the argument inside the root must be larger than or equal to zero. That is:

$2-x\geq 0\\2\geq x\\x\leq 2$

So, this condition must be satisfied by the x-values of the domain.

Then we have the binomial in the denominator, which in order to be defined needs to be different from zero. Notice that the only x-value that could cause problems (render zero) is:

$x-1=0\\x=1$