All categories

3 years ago

In how many ways can Tim draw a red or a green marble from a jar containing 20 red marbles and 4 green marbles?

Mathematics

2 answers:

kondaur [170]3 years ago

7 0

I believe the answer is 24

oksano4ka [1.4K]3 years ago

4 0

The number of ways we can choose from n objects picking r at a time is given by the formula;

$nC_r = \frac{n ! }{(n - r) ! r ! }$

Number of ways Tim can draw 1 red marble from 20 red marbles is

$20C_1 = \frac{20! }{(20 - 1) ! 1 ! } = \frac{20 \times 19 ! }{19 ! 1! } = 20$

Number of ways Tim can draw 1 green marble from 4 green marbles is

$4C_1 = \frac{4! }{(4 - 1) ! 1 ! } = \frac{4 \times 3 ! }{3 ! 1! } = 4$

Number of ways Tim can draw a red or green marble is

$20C_1 +4C_1 = 20 + 4 = 24$

Hence the correct answer is 24.

You might be interested in

Function f is defined by f(x)=2x-7 find f(3)

Zina [86]

Answer:

(

)

−

Step-by-step explanation:

7 0

4 years ago

In a house, there is a small statue in the shape of a cylinder. The cylinder has a diameter of 8 inches and a height of h inches

Kitty [74]

The equation that best represent the volume of the cylinder in cubic inches is V = π4²h.

<h3>What is a cylinder?</h3>

A cylinder is simply a 3-dimensional shape having two parallel circular bases joined by a curved surface.

The volume of a cylinder is expressed as;

V = π × r² × h

Where r is radius of the circular base, h is height and π is constant pi.

Given that;

Diameter d = 8in
Radius r = d/2 = 8in/2 = 4in
Height = h
Volume of the cylinder V = ?

V = π × r² × h

V = π × (4)² × h

V = π4²h

Therefore, the equation that best represent the volume of the cylinder in cubic inches is V = π4²h.

Learn more on volume of cylinder here: brainly.com/question/16788902

#SPJ1

3 0

2 years ago

The asymptote of the function f(x) = 3x + 1 – 2 is ___. Its y-intercept is ___.

Svetradugi [14.3K]

The only way this could have an asymptote, either horizontal, vertical, or oblique is if it is a rational function, proper or improper. This function is linear:
y=3x-1 is what it simplifies to and it is a straight line through the y axis at -1 and has a positive slope of 3. There are no asymptotes here. But the y intercept, like I stated, is (0, -1)

8 0

3 years ago

Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x

soldier1979 [14.2K]

Answer:

(a) The function is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is $x = -\frac{1}{2}$

(d) The function is concave upward on $\left(- \frac{1}{2}, \infty\right)$ and concave downward on $\left(-\infty, - \frac{1}{2}\right)$

Step-by-step explanation:

(a) To find the intervals where $f(x) = 2x^3 + 3x^2 -180x$ is increasing or decreasing you must:

1. Differentiate the function

$\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180$

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

$f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6$

These points divide the number line into three intervals:

$(-\infty,-6)$ , $(-6,5)$ , and $(5, \infty)$

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

$\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right$

Therefore f(x) is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.

f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

$f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6$