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Scilla [17]
3 years ago
10

Brayden's horse gained 9 pounds over a 3 week

Mathematics
2 answers:
larisa [96]3 years ago
7 0

Answer:

about 27 + pounds

Step-by-step explanation:

mr_godi [17]3 years ago
5 0
30 pounds after 10 weeks
9/3=3
3*10=30
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A student knows the height of a pyramid and the area of its base. What should the student do to find the volume of the pyramid
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Answer:

Multiply the area of the base by the height and divide by 3.

Step-by-step explanation:

V = (1/3)Bh

where B = area of the base, and

h = height

Answer: Multiply the area of the base by the height and divide by 3.

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Find the midpoint of the line segment joining the points R(-5,3) and S(1.6)​
Naddik [55]

Answer:

(-2, 4.5)

Step-by-step explanation:

The midpoint formula is (\frac{x_{1} +x_{2}}{2} ,\frac{y_{1} +y_{2}}{2})

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Your answer will be (-2, 4.5)

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4 years ago
Can someone please help me with this
Artist 52 [7]

9514 1404 393

Answer:

  21.  D

  22.  C

Step-by-step explanation:

21. The expansion of the given expression is ...

  \displaystyle -\frac{1}{2}\left(-\frac{3}{2}x+6x+1\right)-3x=\frac{3}{4}x-3x-\frac{1}{2}-3x\\\\=\left(\frac{3}{4}-3-3\right)x-\frac{1}{2}=\boxed{-5\frac{1}{4}x-\frac{1}{2}}

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22. The least likely team to make the championship game is the one with the lowest probability.

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3 years ago
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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.

<h3>A)</h3>

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}.

<h3>B)</h3>

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.

<h3>C)</h3>

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.

Probability:

\displaystyle \frac{560}{1540}= \frac{4}{11}.

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4 years ago
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