3 years ago

The store bought a table for $200, and sold it for $350. What percentage was the markup? %

Mathematics

1 answer:

Zanzabum3 years ago

8 0

Answer: 175% or 75% more in price

Step-by-step explanation:

You might be interested in

Alison is playing a video game. At the end of each level, the player is given either a bag of gold or a magic wand.

Shkiper50 [21]

Outcome Bag of Gold Magic Wand

Relative frequency 0.32 0.68

The relative frequency of getting a bag of gold is .......... reasonably close

.32 is close to 30% so

Alison's claim about the theoretical probability is likely to be 2............true

Further, this means that the theoretical probability of getting a magic wand is most likely 3............1 - 30% = 70%

5 0

3 years ago

How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

Integration by parts will help here. Letting $u=\arctan x$ and $\mathrm dv=\mathrm dx$ , you end up with $\mathrm du=\dfrac{\mathrm dx}{1+x^2}$ and $v=x$ . Now

$\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du$
$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$