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ra1l [238]
3 years ago
15

a professor must randomly select 4 students to participate in a mock debate. There are 20 students in his class. In how many dif

ferent ways can these students be selected, if the number of selections does not matter?
Mathematics
2 answers:
Feliz [49]3 years ago
6 0
If the order doesn't matter the it's
C(20,4)=\dfrac{20!}{4!16!}=\dfrac{17\cdot18\cdot19\cdot20}{2\cdot3\cdot4}=4845
Grace [21]3 years ago
3 0

Answer:

Different ways can these students be selected = 4845

Step-by-step explanation:

Number of combinations possible from n number of population if r people are selected is ^nC_r

Here n = 20 and r = 4

Different ways can these students be selected = ^{20}C_4

^{20}C_4=\frac{20\times 19\times 18\times 17}{1\times 2\times 3\times 4}=4845

Different ways can these students be selected = 4845

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2 years ago
Find the volume of the solid by rotating the region bounded by y=x^3, y=8, and x=0 about the y axis.
dimulka [17.4K]

Answer:

Solution : Volume = 96/5π

Step-by-step explanation:

If we slice at an arbitrary height y, we get a circular disk with radius x, where x = y^(1/3). So the area of a cross section through y should be:

A(y) = πx^2 = π(y^(1/3))^2 = πy^(2/3)

And now since the solid lies between y = 0, and y = 8, it's volume should be:

V =  ∫⁸₀  A(y)dy (in other words ∫ A(y)dy on the interval [0 to 8])

=> π ∫⁸₀ y^(2/3)dy

=> π[3/5 * y^(5/3)]⁸₀

=> 3/5π(³√8)⁵

=> 3/5π2^5

=> 96/5π ✓

7 0
3 years ago
Given f(x) = (lnx)^3 find the line tangent to f at x = 3
kirill [66]
Explanation

We must the tangent line at x = 3 of the function:

f(x)=(\ln x)^3.

The tangent line is given by:

y=m*(x-h)+k.

Where:

• m is the slope of the tangent line of f(x) at x = h,

,

• k = f(h) is the value of the function at x = h.

In this case, we have h = 3.

1) First, we compute the derivative of f(x):

f^{\prime}(x)=\frac{d}{dx}((\ln x)^3)=3*(\ln x)^2*\frac{d}{dx}(\ln x)=3*(\ln x)^2*\frac{1}{x}=\frac{3(\ln x)^2}{x}.

2) By evaluating the result of f'(x) at x = h = 3, we get:

m=f^{\prime}(3)=\frac{3}{3}*(\ln3)^2=(\ln3)^2.

3) The value of k is:

k=f(3)=(\ln3)^3

4) Replacing the values of m, h and k in the general equation of the tangent line, we get:

y=(\ln3)^2*(x-3)+(\ln3)^3.

Plotting the function f(x) and the tangent line we verify that our result is correct:

Answer

The equation of the tangent line to f(x) and x = 3 is:

y=(\ln3)^2*(x-3)+(\ln3)^3

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