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3 years ago

17 20 13 5 30 13 6 15 median

Mathematics

1 answer:

fiasKO [112]3 years ago

5 0

The median is 14.......

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hellllpp, with both number 1 and 2, first one to answer gets brainliest, be sure of your answer please!

Fed [463]

Answer:pretty sure 1 is 23 and 2 is 211200

Step-by-step explanation:

1. 92/4 is 23

2. Just used a calculator on

4 0

2 years ago

Men absolute deviation of 14, 8, 7,6,5,10,11,8,8,6

maks197457 [2]

I don't know much about this but as far as I can tell the mean is about 8, that would make the deviation value of about 2

4 0

3 years ago

It's urgent I need help quick pls help me

Stella [2.4K]

Answer:

Step-by-step explanation:

16 - 4 $\sqrt{5}$ + 4 $\sqrt{5}$ - 5

= 16 - 5

= 11

5 0

3 years ago

If K is the midpoint of JL, JK = 9x -1 and KL = 2x +27, find JL

Liula [17]

Answer:

JL = 70

Step-by-step explanation:

Since K is the midpoint of JL then JK = KL and

JL = JK + KL = 9x - 1 + 2x + 27 = 11x + 26

solve for x using JK = KL

9x - 1 = 2x + 27 ( subtract 2x from both sides )

7x - 1 = 27 ( add 1 to both sides )

7x = 28 ( divide both sides by 7 )

x = 4, hence

JL = 11x + 26 = (11 × 4 ) + 26 = 44 + 26 = 70

3 0

3 years ago

Read 2 more answers

Lim x approaches 0 (1+2x)3/sinx

jok3333 [9.3K]

Interpreting your expression as

$\dfrac{3(1+2x)}{\sin(x)}$

when $x$ approaches zero, the numerator approaches 3:

$3(1+2x) \to 3(1+2\cdot 0) = 3(1+0) = 3\cdot 1 = 3$

The denominator approaches 0, because $\sin(0)=0$

Moreover, we have

$\displaystyle \lim_{x\to 0^-} \sin(x) = 0^-,\quad \displaystyle \lim_{x\to 0^+} \sin(x) = 0^+$

So, the limit does not exist, because left and right limits are different:

$\displaystyle \lim_{x\to 0^-} \dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^-} = -\infty,\quad \displaystyle \lim_{x\to 0^+}\dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^+} = +\infty$

8 0

3 years ago