All categories

1 year ago

How many different sequences of 3 playing cards (from a single 52-card deck) exist?

Mathematics

1 answer:

hichkok12 [17]1 year ago

4 0

There 1,32,600 possibilities to withdrawn 3 different sequence of playing cards from 52 cards.

<h3>What is Permutations?</h3>

A permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.

Here, total number of cards = 52

Number of Withdrawn cards = 3

Possibilities of different sequences of 3 playing cards = 52 X 51 X 50

= 132600

Thus, there 1,32,600 possibilities to withdrawn 3 different sequence of playing cards from 52 cards.

Learn more about Permutations from:

brainly.com/question/1216161

#SPJ1

You might be interested in

Simplify the equation 4x+3y=2x

nikdorinn [45]

Answer:

2x=3y or x= -3y/2 and y =2/-3y

Step-by-step explanation:

4x-2x=-3y

2x=-3y

4 0

3 years ago

Read 2 more answers

Least common denominator?

Over [174]

Answer:

$\large\boxed{(s-2)(s+2)(s-3)=(s^2-4)(s-3)=s^3-3s^2-4s+12}$

Step-by-step explanation:

$s^2-4=s^2-2^2=(s-2)(s+2)\\\\\text{used}\ a^2-b^2=(a-b)(a+b)\\\\s^2-5s+6=s^2-2s-3s+6=s(s-2)-3(s-2)=(s-2)(s-3)\\\\\text{Therefore}\ LCD(s^2-4,\ s^2-5s+6)=(s-2)(s+2)(s-3)$

4 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

2 years ago

It takes Mario and Jessica 15 minutes to set up 5 tables. If Mario and Jessica work at a constant rate, which best describes a p

AfilCa [17]

Answer:

3 minutes per table

Step-by-step explanation:

15 minutes = 5 tables

to get how much it will be per table, divided both sides by 5

since 5 / x = 1 and x would be 5

15 / 5 minutes = 5 / 5 tables

3 minutes = 1 table

5 0

2 years ago

BACTROBAN ointment contains 2% w/w mupirocin. How many grams of a polyethylene glycol ointment base must be mixed with the conte

lyudmila [28]

Answer:

66g of polyethylene ointment base!

6 0

3 years ago