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3 years ago

I don’t know how to do this problem. PLEASE HELP ME!

Mathematics

1 answer:

Usimov [2.4K]3 years ago

8 0

Answer:

I really don't know what's the answer

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2x+2y=8 pleaseee help!!

Charra [1.4K]

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3 years ago

What is the y-intercept of the graph of 4y -1=3?

inn [45]

Answer:

(0,1)

Step-by-step explanation:

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4 years ago

Read 2 more answers

Please help me. Need it.

Lunna [17]

Hey There!
win 10 lose 12
win 15 lose 18

5×2=10, 6×2=12
5×3=15, 6×3=18

Hope This Helps!!!

4 0

3 years ago

Plssss help

Andre45 [30]

Answer:

A) $\displaystyle \frac{1}{77}$ .
B) $\displaystyle \frac{12}{77}$ .
C) $\displaystyle \frac{4}{11}$ .

Step-by-step explanation:

All marbles here are identical. Also, the question isn't concerned about the order in which the marbles are drawn. Thus, all calculations here shall be combinations rather than permutations.

How many ways to choose three out of six identical red marbles without replacement?

$\displaystyle _6C_3 = c(6, 3) = {6\choose 3} = 20$ .

Note that these three expressions are equivalent. They all represent the number of ways to choose 3 out of 6 identical items without replacement.

How many ways to choose three out of all the 6 + 10 + 6 = 22 marbles?

$\displaystyle _{22} C_{3} = 1540$ .

The probability of choosing three red marbles out of these 22 marbles will be:

$\displaystyle \frac{\text{Number of ways for choosing three out of six red marbles}}{\text{Number of ways to choose three out of 22 marbles}} = \frac{20}{1540} = \frac{1}{77}$ .

How many ways to choose two out of six identical red marbles without replacement?

$\displaystyle _6 C_2 = 15$ .

How many ways to choose one out of 10 + 6 = 16 non-red marbles?

$_{16} C_1=16$ .

Choosing two red marbles does not influence the number of ways of choosing a non-red marble. Both event happen and are independent of each other. Apply the product rule to find the number of ways of choosing two red marbles and one non-red marble out of the pile of 22.

$_6 C_2 \cdot _{16} C_1= 240$ .

Probability:

$\displaystyle \frac{240}{1540} = \frac{12}{77}$ .

Double check that the order doesn't matter here.

None of the marbles are red. In other words, all three marbles are chosen out of a pile of 10 + 6 = 16 white and blue marbles. Number of ways to do so:

$_{16} C_{3} = 560$ .