${ \tt{first \: rotation = 360 \degree}} \\ { \tt{second \: rotation = 360 \degree}} \\ { \tt{third \: rotation = 90 \degree + (90 - 60) \degree}} \\ { \tt{third \: rotation = 120 \degree}} \\ \\ { \bf{angle}} = { \tt{ {1}^{st} \: rotation + {2}^{nd} \: rotation + {3}^{rd} \: rotation}} \\ = 360 \degree + 360\degree + 120\degree \\ = { \tt{840\degree}}$

7 0

2 years ago

Evaluate the definite integral. 1/4 csc(2πt) cot(2πt) dt 1/12

Afina-wow [57]

Hey there, hope I can help!

$\frac{1}{4}\csc \left(2\pi t\right)\cot \left(2\pi t\right)dt\frac{1}{12}$

$\mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}\cdot \frac{d}{e}=\frac{a\:\cdot \:b\:\cdot \:d}{c\:\cdot \:e} \ \textgreater \ \frac{1\cdot \:1\cdot \:dt}{4\cdot \:12}\csc \left(2\pi t\right)\cot \left(2\pi t\right)$

$\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \ \frac{dt}{4\cdot \:12}\csc \left(2\pi t\right)\cot \left(2\pi t\right)$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:12=48 \ \textgreater \ \frac{dt}{48}\csc \left(2\pi t\right)\cot \left(2\pi t\right)$

$\mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{dt\csc \left(2\pi t\right)\cot \left(2\pi t\right)}{48}$

Hope this helps!

5 0

3 years ago

Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red

Neporo4naja [7]

Answer:

Multiple answers

Step-by-step explanation:

The original urns have:

Urn 1 = 2 red + 4 white = 6 chips
Urn 2 = 3 red + 1 white = 4 chips

We take one chip from the first urn, so we have:

The probability of take a red one is : $\frac{1}{3}$ (2 red from 6 chips(2/6=1/2))

For a white one is: $\frac{2}{3}$ (4 white from 6 chips(4/6=(2/3))

Then we put this chip into the second urn:

We have two possible cases:

First if the chip we got from the first urn was white. The urn 2 now has 3 red + 2 whites = 5 chips
Second if the chip we got from the first urn was red. The urn two now has 4 red + 1 white = 5 chips

If we select a chip from the urn two:

In the first case the probability of taking a white one is of: $\frac{2}{5}$ = 40% ( 2 whites of 5 chips)
In the second case the probability of taking a white one is of: $\frac{1}{5}$ = 20% ( 1 whites of 5 chips)

This problem is a dependent event because the final result depends of the first chip we got from the urn 1.

For the fist case we multiply :

$\frac{4}{6}$ x $\frac{2}{5}$ = $\frac{4}{15}$ = 26.66% ( $\frac{4}{6}$ the probability of taking a white chip from the urn 1, $\frac{2}{5}$ the probability of taking a white chip from urn two)

For the second case we multiply:

$\frac{1}{3}$ x $\frac{1}{5}$ = $\frac{1}{30}$ = .06% ( $\frac{1}{3}$ the probability of taking a red chip from the urn 1, $\frac{1}{5}$ the probability of taking a white chip from the urn two)

8 0

3 years ago