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3 years ago

3926%13. I got 309 r9. What is it and what did I do wrong.

Mathematics

1 answer:

kondaur [170]3 years ago

6 0

3926/13 =302 but I'm not 100% sure where you went wrong do you have any work?

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I really need this answer to be correct

Fed [463]

It is unchanged because the same number would still be in the middle and there would still be the same amount of numbers

if
11,15,21,22,23,27,30 before 22 is the median

if
11,15,21,22,23,27,30 after 22 is still the median

The median is unchanged

4 0

4 years ago

There are 12,000 people attending a concert. They announce that 20% of the people that purchased tickets did not come to the con

Marysya12 [62]

Answer:

15,000

Step-by-step explanation:

20% of 15,000 is 3,000. 15,000-3,000=12,000

8 0

3 years ago

Simplify your answer

Lelechka [254]

Answer:

Step-by-step explanation:

If 2u-6=4, let's solve for u.

Add 6 to both sides

2u-6+6 = 4+6

Simplify

2u = 10

Divide both sides by 2

u = 5

Now, we can solve u-5

5-5 = 0

8 0

2 years ago

Read 2 more answers

In how many different ways can a red, blue, and green marker be arranged in a tray?

grigory [225]

Answer:

6 ways.

There are one red, one blue and one green maker. So there are 3 markers they are in 3! = 3×2 = 6 ways.

Please mark brainliest! <3

7 0

3 years ago

What is the product?

fredd [130]

Answer:

$\frac{4k + 2}{k^{2}-4 }$ × $\frac{k-2}{2k+1}$ = $\frac{2}{k + 2}$

Step-by-step explanation:

$\frac{4k + 2}{k^{2}-4 }$ × $\frac{k-2}{2k+1}$

To solve the above, we need to follow the steps below;

4k+2 can be factorize, so that;

4k +2 = 2 (2k + 1)

k² - 4 can also be be expanded, so that;

k² - 4 = (k-2)(k+2)

Lets replace 4k +2 by 2 (2k + 1)

and

k² - 4 by (k-2)(k+2) in the expression given

$\frac{4k + 2}{k^{2}-4 }$ × $\frac{k-2}{2k+1}$

$\frac{2(2k+ 1)}{(k-2)(k+2)}$ × $\frac{k-2}{2k+1}$

(2k+1) at the numerator will cancel-out (2k+1) at the denominator, also (k-2) at the numerator will cancel-out (k-2) at the denominator,

So our expression becomes;

$\frac{2}{k + 2}$

Therefore, $\frac{4k + 2}{k^{2}-4 }$ × $\frac{k-2}{2k+1}$ = $\frac{2}{k + 2}$

3 0

3 years ago

Read 2 more answers