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GrogVix [38]
3 years ago
7

Elimination3x +2y=93x ty=01​

Mathematics
1 answer:
muminat3 years ago
8 0

Answer:

y = 9

x = -3

Step-by-step explanation:

Given

3x + 2y = 9

3x + y = 0

Required

Solve by elimination

To do this, we subtract both equations (to eliminate x)

3x + 2y = 9

- 3x + y = 0

--------------------

3x - 3x + 2y - y = 9-0

y = 9

Substitute y = 9 in 3x + y = 0

3x + 9 = 0

Solve for 3x

3x =- 9

Solve for x

x = -9/3

x = -3

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Normally 7 envelopes but on rare occasions, there may be more due to If the packages are filled with Envelopes. Also depending on if the 7 packages are boxes of envelopes, and in that case, 140 envelopes times 7, which would be 980 envelopes in the 7 packages.      140 * 7 = 980.
5 0
4 years ago
We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely. (a) Find the probability th
Verizon [17]

Answer:

a) There is a 16.6% probability that doubles were rolled.

b) Given that the roll resulted in a sum of 6 or less, there is a 20% probability that doubles were rolled.

c) There is a 30.55% probability that at least one die is a 1.

Step-by-step explanation:

The probability formula is the number of desired outcomes divided by the number of total outcomes.

(a) Find the probability that doubles were rolled. (i.e. the two outcomes are equal)

There are six desired outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

There are 36 total outcomes. So:

P = \frac{6}{36} = 0.166

There is a 16.6% probability that doubles were rolled.

(b) Given that the roll resulted in a sum of 6 or less, find the conditional probability that doubles were rolled.

The following rolls result in a sum of 5 or less:

(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1)

So, there are 15 total outcomes.

3 of them are doubled: (1,1), (2,2), (3,3). So

P = \frac{3}{15} = 0.2

Given that the roll resulted in a sum of 6 or less, there is a 20% probability that doubles were rolled.

(c) Find the probability that at least one die is a 1

The following outcomes have at least one die that is 1:

(1,1), (1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1)

So, there are 11 desired outcomes out of 36.

P = \frac{11}{36} = 0.3055

There is a 30.55% probability that at least one die is a 1.

6 0
4 years ago
Please help I wanna get an a :(
kramer

Answer:

Graph C

Step-by-step explanation:

Point T is located on Graph C at T(1, 7). To help, (x, y) = count over however many spaces <u>horizontally</u>, then count however many spaces <u>vertically</u>.

<em>It will always be this way.</em>

4 0
3 years ago
Line Symmetry and Reflections help?
QveST [7]

5.(8)

6. (D)

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8 0
4 years ago
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Edgar ran e meters per second and mathieu ran m meters per second the boys ran for t seconds
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Te, tm, m-e, t. The last one is correct
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