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3 years ago

From a pack of 9 cards numbered 1-9, three cards are drawn at random and laid on a table from left to right.

1 answer:

4 0

Answer: 1/6
Explanation: Since we need to draw three cards from a pack of 9 and the digits require an order, then our sample space becomes: 9P3.
Now, we don't care about picking an order when picking three cards to analyse, as there is only 1 way in arranging the three cards in descending order.

Thus, we get:

$\frac{^{9}C_3}{^{9}P_3} = \frac{1}{6}$

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What is the difference between the cube root of a number and the product of x and y

olga nikolaevna [1]

A number y plus the cube monomial expressions and cube roots of whole numbers

7 0

3 years ago

On a coordinate plane, a line is drawn from point A to point B. Point A is at (2, negative 3) and point B is at (negative 4, 9).

zimovet [89]

Answer: (0,1)

Step-by-step explanation:

If $(x_1, y_1)$ and $(x_2, y_2)$ are two point son a coordinate plane and (x,y) dividing it in a ratio of m: n.

Then , the coordinates of (x,y) is given by :-

$x=\dfrac{nx_1+mx_2}{m+n}$

$y=\dfrac{ny_1+my_2}{m+n}$

Given : On a coordinate plane, a line is drawn from point A to point B. Point A is at (2, - 3) and point B is at (- 4, 9).

Then , the x- and y- coordinates of point E, which partitions the directed line segment from A to B into a ratio of 1:2 :

$x=\dfrac{2(2)+1(-4)}{1+2}=0$

$y=\dfrac{2(-3)+1(9)}{1+2=1}$

Hence, the x- and y- coordinates of point E = (0,1)

3 0

3 years ago

Read 2 more answers

The sides of ∠A are tangent to circle k(O) with radius r. Find: r, if OA=14 dm, m∠A=90°.

Alex_Xolod [135]

Answer:

r = √98

Step-by-step explanation:

angle A and the diameter form an isosceles right triangle, with OA as the hypotenuse and r as the other sides. You can then make and solve an equation from the Pythagorean Theorem:

r^2 + r^2 = 14^2

2r^2 = 14^2

2r^2 = 196

r^2 = 98

r = √98

6 0

3 years ago

Read 2 more answers

Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

4 0

3 years ago

You buy a car for $25,000. It depreciates at the rate of 23% per year.

Juliette [100K]

The worth of the car in 10 years is $1,831.67 using an exponential equation approach.

What is an exponential equation?

An exponential equation is the one with exponents such X^3(the 3 is the exponent)

The exponential equation required here is the one where the future value would be lower than current value because the car reduces in value year-in-year-out.

FV=PV*(1-r)^N

FV=future worth of the car

PV=today's value=$25,000

r=depreciation rate=-23%

N=number of years=10

The fact that r is negative means the car is depreciating not appreciating.

FV=$25,000*(1-23%)^10

FV=$1,831.67

Find further explanation on exponential equation below in the link:

brainly.com/question/11832081

#SPJ1

7 0

2 years ago