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solmaris [256]
3 years ago
15

If there are three black, four white, two blue, and four gray socks in a drawer, what would be the probability of picking a blue

sock? Round your answer to the nearest tenth.
Mathematics
1 answer:
Alborosie3 years ago
8 0

Answer:

3/13 which is 0.23 as a decimal

Step-by-step explanation:

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HELP PLEASE<br> I WILL GIVE BRAINLIEST
Olegator [25]

Answer:

<h2>B</h2>

I hoped this helped :))

6 0
3 years ago
Read 2 more answers
8p + 2q = -16<br> 2p - q=2
yKpoI14uk [10]

Answer:

p = -1  q = -4

Step-by-step explanation:

a system of eq and solve for p and q ???  can do  :)

Eq. 1)  8p + 2q = - 16

Eq. 2)  2p - q = 2

use Eq .2 and solve for q

2p - 2 = q

plug into Eq.1 with q

8p +2(2p - 2) = - 16

8p +4p -4 = -16

12p = - 12

p = -1

plug -1 into Eq. 1 for p and solve for q

8(-1) + 2q = - 16

-8 + 2q = - 16

2q = -8

q = -4

8 0
3 years ago
The numbers of inches of snow per month in Portland, Maine, for a period of twelve months, are:
Gnom [1K]
The range is 19-0=19 so therefore your answer is C.
The range is the largest number subtracted by the smallest number.
8 0
3 years ago
Read 2 more answers
You picked one marble out of a bag that contains 6 red marbles, 5 white marbles, and 3 blue marbles. Find p(red and blue).
Tcecarenko [31]
Answer: p=9
explanation: 6 reds + 3 blues = 9.
6 0
3 years ago
HELP PRECALC DO NOT UNDERSTAND WILL GIVE BRAINLIEST
Olegator [25]

Answer:

  the lower right matrix is the third correct choice

Step-by-step explanation:

Your problem statement shows that you have correctly selected the matrices representing the initial problem setup (middle left) and the problem solution (middle right).

Of the remaining matrices, the upper left is an incorrect setup, and the lower left is an incorrect solution matrix.

__

We notice that in the remaining matrices on the right that the (2,3) term is 0, and the (3,2) and (3,3) terms are both 1.

The easiest way to get a 0 in the 3rd column of row 2 is to add the first row to the second. When you do that, you get ...

  \left[\begin{array}{ccc|c}1&1&1&29000\\1+2&1-3&1-1&1000(29+1)\\0&0.15&0.15&2100\end{array}\right] =\left[\begin{array}{ccc|c}1&1&1&29000\\3&-2&0&30000\\0&0.15&0.15&2100\end{array}\right]

Already, we see that the second row matches that in the lower right matrix.

The easiest way to get 1's in the last row is to divide that row by 0.15. When we do that, the (3,4) entry becomes 2100/0.15 = 14000, matching exactly the lower right matrix.

The correct choices here are the two you have selected, and <em>the lower right matrix</em>.

3 0
3 years ago
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