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lara31 [8.8K]
3 years ago
6

How many different committees can be formed from 8 teachers and 48 students if the committee consists of 3 teachers and 4 ​stude

nts?
Mathematics
2 answers:
valkas [14]3 years ago
6 0
As many as you'd like
drek231 [11]3 years ago
6 0
Since we don't care about the order in which we select people, we will use the combinatoric approach, rather than permutations.
Since we want to select 3 teachers from a total of 8, this is the notation to select 3 teachers from 8 (objects).

Teachers : ^{8}C_3
Similarly, students : ^{48}C_{4}

Now, since we want both at the same time, we will need to multiply both to yield your final answer: ^{8}C_3 \cdot ^{48}C_{4} = 10,896,480 different ways.
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Math please help 10 points!
gizmo_the_mogwai [7]

Answer:


Step-by-step explanation:

average= \frac{sum of scores}{number of tests}

80= \frac{79+81+77+x}{4}

80 x 4 = 79+81+77+x

320=237  + x

320-237 = x

83=x

so Marianne must score 83 points or higher on her last test

8 0
3 years ago
What is an equation for the linear function whose graph contains the points (−1, −2) and (3, 10)
sladkih [1.3K]
The equation of a line starting from two points is:
y-y_1=\frac{y_2-y_1}{x_2-x_1} \cdot (x-x_1)

From the first point you get: x1 = -1, y1 = -2
From the second point you get: x2 = 3, y2 = 10

Replace x1, y1, x2, y2 in the equation of the line and you get:
y+2=\frac{10+2}{3+1} \cdot (x+1)
y+2=\frac{12}{4} \cdot (x+1)
y+2=3 \cdot (x+1)

From this you get the equation of your line:
y(x)=3x+1




4 0
3 years ago
Read 2 more answers
In a circle with radius 9, an angle measuring pi over 3 radians intercepts an arc. Find the length of the arc in simplest form
BlackZzzverrR [31]

Answer:

6

Step-by-step explanation:

6 0
3 years ago
Fill in the missing terms.
san4es73 [151]

a) 2

b) 2

c) 6

d) 15

e) 2

f) 3

6 0
3 years ago
Solve the following system of equations. (Hint: Use the quadratic formula.) f(x) = 2x² 3x g(x)=-3x² + 20 (0.-10) and (1, 17) (-2
jasenka [17]

The solution of the system of equation is the intersection point of the two quadratic equations, so we need to equate both equations, that is,

2x^2-3x-10=-3x^2+20

So, by moving the term -3x^3+20 to the left hand side, we have

5x^2-3x-30=0

Then, in order to solve this equation, we can apply the quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

In our case, a=5, b=-3 and c=-30. So we get

x=\frac{3\pm\sqrt{(-3)^2-4(5)(-30)}}{2(5)}

which gives

\begin{gathered} x=2.76779 \\ and \\ x=-2.16779 \end{gathered}

By substituting these points into one of the functions, we have

f(2.76779)=-2.982

and

f(-2.16779)=5.902

Then, by rounding these numbers to the nearest tenth, we have the following points:

\begin{gathered} (2.8,-3.0) \\ and \\ (-2.2,5.9) \end{gathered}

Therefore, the answer is the last option

4 0
1 year ago
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