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Step2247 [10]
3 years ago
11

find the number of ways in which six teachers can be assigned to four sections of introductory mathematics if no teacher must be

assigned to more than one section?
Mathematics
2 answers:
SpyIntel [72]3 years ago
7 0
So 4 sections
6 options for first section
5 options for second (since one was used for first)
4 options
3 options
mutiply
6 times 5 times 4 times 3=360

answer is 360 ways
irina1246 [14]3 years ago
7 0

The first section can be taught by any one of
the 6 teachers.                               For each one . . .
The 2nd section can be taught by any one of
the remaining 5 teachers.              For each one . . .
The 3rd section can be taught by any one of
the remaining 4 teachers.              For each one . . .
The 4th section can be taught by any one of
the remaining 3 teachers.

So the number of possible teams of teachers for all four
math sections is
                
                   (6 x 5 x 4 x 3)  =  360 different teams.

Notice that whichever team is selected, there will be two teachers
who have no class to teach during the introductory math hour.

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Help please! Geometry word problem!
mr Goodwill [35]

By applying <em>reflection</em> theory and constructing a <em>geometric</em> system of two <em>proportional right</em> triangles, the height of the stainless steel globe is approximately 140 ft.

<h3>How to estimate the height of the stainless steel globe</h3>

By physics we know that both the angle of incidence and the angle of reflection are same. Thus, we have a <em>geometric</em> system formed by two <em>proportional right</em> triangles:

5.6 ft / 4 ft = h / 100 ft

h = (5.6 ft × 100 ft) / 4ft

h = 140 ft

By applying <em>reflection</em> theory and constructing a <em>geometric</em> system of two <em>proportional right</em> triangles, the height of the stainless steel globe is approximately 140 ft.

To learn more on geometry: brainly.com/question/16836548

#SPJ1

8 0
1 year ago
QUICK HELP PLEASE WILL GIVE BRAINLIEST
Free_Kalibri [48]

is it multiple choice


7 0
2 years ago
How many solutions does the system of equations below have?
soldier1979 [14.2K]

Answer:

One solution                    

Step-by-step explanation:

5x + y = 8

15x + 15y = 14

Lets solve using substitution, first we need to turn "5x + = 8" into "y = mx + b" or slope - intercept form

So we solve for "y" in the equation "5x + y = 8"

5x + y = 8

Step 1: Subtract 5x from both sides.

5x + y − 5x = 8 − 5x

Step 2: 5x subtracted by 5x cancel out and "8 - 5x" are flipped

y = −5x + 8

Now we can solve using substitution:

We substitute "-5x + 8" into the equation "15x + 15y = 14" for y

So it would look like this:

15x + 15(-5x + 8) = 14

Now we just solve for x

15x + (15)(−5x) + (15)(8) = 14(Distribute)

15x − 75x + 120 = 14

(15x − 75x) + (120) = 14(Combine Like Terms)

−60x + 120 = 14

Step 2: Subtract 120 from both sides.

−60x + 120 − 120 = 14 − 120

−60x = −106

Divide both sides by -60

\dfrac{ -60x  }{ -60  }   =   \dfrac{ -106  }{ -60  }

Simplify

x =   \dfrac{ 53  }{ 30  }

Now that we know the value of x, we can solve for y in any of the equations, but let's use the equation "y = −5x + 8"

\mathrm{So\:it\:would\:look\:like\:this:\ y =  -5 \left(  \dfrac{ 53  }{ 30  }    \right)  +8}

\mathrm{Now\:lets\:solve\:for\:"y"\:then}

y =  -5 \left(  \dfrac{ 53  }{ 30  }    \right)  +8}

\mathrm{Express\: -5 \times   \dfrac{ 53  }{ 30  }\:as\:a\:single\:fraction}

y =   \dfrac{ -5 \times  53  }{ 30  }  +8

\mathrm{Multiply\:-5 \:and\:53\:to\:get\:-265 }

y =   \dfrac{ -265  }{ 30  }  +8

\mathrm{Simplify\:  \dfrac{ -265  }{ 30  }    \:,by\:dividing\:both\:-265\:and\:30\:by\:5} }

y =   \dfrac{ -265 \div  5  }{ 30 \div  5  }  +8

\mathrm{Simplify}

y =  - \dfrac{ 53  }{ 6  }  +8

\mathrm{Turn\:8\:into\:a\:fraction\:that\:has\:the\:same\:denominator\:as\: - \dfrac{ 53  }{ 6  }}

\mathrm{Multiples\:of\:1: \:1,2,3,4,5,6}

\mathrm{Multiples\:of\:6: \:6,12,18,24,30,36,42,48}

\mathrm{Convert\:8\:to\:fraction\:\dfrac{ 48  }{ 6  }}

y =  - \dfrac{ 53  }{ 6  }  + \dfrac{ 48  }{ 6  }

\mathrm{Since\: - \dfrac{ 53  }{ 6  }\:have\:the\:same\:denominator\:,\:add\:them\:by\:adding\:their\:numerators}

y =   \dfrac{ -53+48  }{ 6  }

\mathrm{Add\: -53 \: and\: 48\: to\: get\:  -5}

y =  - \dfrac{ 5  }{ 6  }

\mathrm{The\:solution\:is\:the\:ordered\:pair\:(\dfrac{ 53  }{ 30  }, - \dfrac{ 5  }{ 6  })}

So there is only one solution to the equation.

5 0
2 years ago
Can someone plzz help​
tankabanditka [31]

Answer:

it's 12 because

2×6=12

3×4=12

4×3=12

and all of them have twelve

6 0
3 years ago
Suppose you drive a car 392 mi on a tank of gas. The tank holds 14 gallons. The number of miles traveled vaties directly with th
Serhud [2]

Preliminary Problem

Givens

Miles = m = 392 miles

Gallons =  G = 14 gallons

Miles per gallon = mpg

Formula

miles = miles per gallon * number of gallons.

Solve

392 = mpg * 14                         Divide by 14

392 / 14 = mpg

28 = mpg  which is really quite good.  Our vehicle gets 44 mpg but it's a 4 cylinder. It's meant to get that kind of milage.

Part A

Let x = the number of gallons used

Let y = the number of miles driven

y = k*x

Part B

y = k* x

k =28 mpg

11700 = 28*x

11700 /28 = x

x = 417.86

If this is incorrect please put some more facts in. The question does not seem complete.


7 0
3 years ago
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