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aksik [14]
3 years ago
10

What is 5C3? A. 35 B. 10 C. 14 D. 28

Mathematics
2 answers:
galben [10]3 years ago
5 0

Answer: B. 10

-5C3 or 5 choose 3 refers to how many combinations are possible from 5 items, taken 3 at a time. To calculate combinations, we will use the formula nCr = n! / r! ... * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time.

-10 is the total number of all possible combinations for choosing 3 elements at a time from 5 distinct elements without considering the order of elements in statistics & probability surveys or experiments. The number of combinations for sample space 5 CHOOSE 3 can also be written as 5C3 in the format of nCr or nCk.

Step-by-step explanation: Hope this help :D

Andreyy893 years ago
4 0

Answer:

10

Step-by-step explanation:

I took the test

Answer: B. 10

-5C3 or 5 choose 3 refers to how many combinations are possible from 5 items, taken 3 at a time. To calculate combinations, we will use the formula nCr = n! / r! ... * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time.

-10 is the total number of all possible combinations for choosing 3 elements at a time from 5 distinct elements without considering the order of elements in statistics & probability surveys or experiments. The number of combinations for sample space 5 CHOOSE 3 can also be written as 5C3 in the format of nCr or nCk.

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faltersainse [42]

Answer:

24

Step-by-step explanation:

120/6=20

300/15=20

360/18=20

480/20=24

Check:

480/24=20

6 0
3 years ago
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Determine the slope and the y intercept .6y+x=6 please help
wel

Answer: the slope is negative  \frac{1}{6} and the y intercept is (0,1)

Step-by-step explanation:

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3 years ago
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katen-ka-za [31]

Answer:12

Step-by-step explanation:

15^2-9^2=12^2

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3 0
3 years ago
Simplify. (write without the absolute value sign): |x+3|, if x=2
iren2701 [21]

Answer:

5

Step-by-step explanation:

|x+3| when x = 2 is |2+3| = |5| = 5

4 0
3 years ago
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How do I solve this algebra two btw
pychu [463]

\bf \begin{cases} x^2+y^2=400\\ \boxed{y}=x-28 \end{cases}\qquad \stackrel{\textit{substituting on the 1st equation}}{x^2+\left( \boxed{x-28} \right)^2=400} \\\\\\ x^2+(x^2-56x+28^2)=400\implies x^2+x^2-56x+28^2=400 \\\\\\ 2x^2-56x+784=400\implies 2x^2-56x+384=0 \\\\\\ 2(x^2-28x+192)=0\implies x^2-28x+192=0 \\\\\\ (x-16)(x-12)=0\implies x = \begin{cases} 16\\ 12 \end{cases}

now that we know what are the x-values, what are the y-values? well, we can just use the 2nd equation, since we know that y = x - 28, then

\bf y = x - 28\implies \stackrel{\textit{when x = 16}}{y = 16 - 28}\implies y = -12 \\\\\\ y = x - 28\implies \stackrel{\textit{when x = 12}}{y = 12 - 28}\implies y = -16 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{x}{16}~~,~~\stackrel{y}{-12})\qquad,\qquad (\stackrel{x}{12}~~,~~\stackrel{y}{-16})~\hfill

7 0
3 years ago
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