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Contact [7]
3 years ago
12

In a high school band there are 11 students in the drumline section. The drumline for a football game consists of 6 students. In

how many ways can the drumline be formed for a football game?
Mathematics
1 answer:
Vanyuwa [196]3 years ago
7 0
If the band consists of 11 students and you have to choose 6 students for<span> the drumline for a football game, then you have to calculate the number of combinations:
</span>
C_{11}^6= \frac{11!}{6!\cdot (11-6)!}= \frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11}{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot1\cdot2\cdot3\cdot4\cdot5}  = \frac{7\cdot8\cdot9\cdot10\cdot11}{2\cdot3\cdot4\cdot5} =462.

Answer: 462 ways to complete the <span>drumline for a football game.</span>
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Will give Brainly lollllllll XD......................
marshall27 [118]

Answer:

it looks like it would just be -1

Step-by-step explanation:

there two -x and two positive x which would cancel out. three positive 1 and four negative 1. -4+3= -1

that's the best I could come up with

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3 years ago
one sixth of the students in a school marching band are in the percussion section, the percussion section has 6 students, how ma
Igoryamba

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36 students

Step-by-step explanation:

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3 years ago
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How do I solve it I’m getting it wrong
k0ka [10]

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Step-by-step explanation:

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2 years ago
onsider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutti
Sholpan [36]

Answer:

V (max)  = 2 ft³

and x side of the base is  x  =  0,5 feet

Step-by-step explanation:  See annex ( two different cubes)

We have a square piece of cardboard of  3 inches wide

Let  x be lenght of side to cut in each corner

Then the base of the (future cube) is    3  -  2x,  and the area is  

( 3 - 2x )²

And  the height   is x  Then volume of the cube as a function of x is:

V(x)  =  ( 3 - 2x )² *x   or       V(x)  =    ( 9 + 4x² - 12x )*x

V(x)  =  4x³  -  12x²  + 9x

Taking derivatives on both sides of the equation

V´(x)  =  12x² - 24x  + 9

V´(x)  =  0       12x² - 24x  + 9  =  0   simplifying   4x² - 8x  + 3  =  0

Second degree equation solving for x

x₁,₂ =  [ 24 ± √( 576) - 432  /24

x₁,₂ =  [24 ±√144 ]/24

x₁,₂ = ( 24 ± 12) /24         x₁  =  1.5 feet        x₂  = 0,5 feet

Of these two values we have to dismiss x₁  because if  x = 1.5 we don´t have a cube ( 0 height )

Then we take x  = 0,5  feet

And

V (max)  =  (2)²*0,5   =  4*0,5

V (max)  = 2 ft³

5 0
3 years ago
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