The origin has an ordered pair of _______

Can you put these number from least to greatest?

Eva8 [605]

The correct order would be:

5/64 x 3, 1/16 x 3, 3/32 x 4, 11/64 x 4, 7/16 x 3, 3/4 x 2, 3/8 x 4, 1 7/8 x 4, 2.25 x 2, 1.5 x 4, 3 3/8 x 3, 3.75 x 3

First we have to take all of the numbers and do the multiplication. It's often easiest to turn them in to decimals so that you have a common form.

3/32 x 4 = 3/8 = .375

3/4 x 2 = 3/2 = 1.5

1 7/8 x 4 = 15/4 = 3.75

2.25 x 2 = 4.5

1.5 x 4 = 6

3/8 x 4 = 3/2 = 1.5

5/64 x 3 = 5/32 = .156

3.75 x 3 = 11.25

1/16 x 3 = 3/16 = .1875

7/16 x 3 = 21/16 = 1.31

3 3/8 x 3 = 81/8 = 10.125

11/64 x 4 = 11/16 = .687

Now we can use those to put in order.

5/64 x 3 = 5/32 = .156

1/16 x 3 = 3/16 = .1875

3/32 x 4 = 3/8 = .375

11/64 x 4 = 11/16 = .687

7/16 x 3 = 21/16 = 1.31

3/4 x 2 = 3/2 = 1.5

3/8 x 4 = 3/2 = 1.5

1 7/8 x 4 = 15/4 = 3.75

2.25 x 2 = 4.5

1.5 x 4 = 6

3 3/8 x 3 = 81/8 = 10.125

3.75 x 3 = 11.25

Which if you are looking for without the extra terms, you can check the answer at the top.

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)