All categories

3 years ago

Solve the problem.

Mathematics

2 answers:

Alika [10]3 years ago

7 0

.00003509

Hope this helps :)

Dafna11 [192]3 years ago

3 0

Answer:

Step-by-step explanation:

Combination question

Generally

nCr = n! / (n-r)!r!

The organizer want to select 5 people from 30 people

Then, the total possible selection is

30C5 = 30! / (30-5)!5!

30C5 = 30•29•28•27•26•25!/25! ×5!

30C5 = 17,100,720 /5•4•3•2

30C5 = 142,506 ways

So, there are 142,506 numbers of ways to selected them.

The group of 5 youngest persons of 30 is UNIQUE such group of 5 persons from 30, under given conditions..

if there are exactly 5 youngest people and if they were all selected, the probability is

5 / 142,506 = 0.000035086

You might be interested in

What is 1,112 rounded to the nearest thousand?

DENIUS [597]

Answer: 1000

because you round down

7 0

3 years ago

Consider the surface z = x 2+4y 2+1. Suppose you are walking on this surface directly above a curve C in the xy-plane, where the

Bess [88]

With $x(t)=\cos t$ and $y(t)=\sin t$ , we have

$z(t)=x(t)^2+4y(t)^2+1$

$z(t)=\cos^2t+4\sin^2t+1$

$z(t)=3\sin^2t+2$

Then $z(t)$ has critical points where

$\dfrac{\mathrm dz}{\mathrm dt}=6\sin t\cos t=3\sin2t=0$

$\implies\sin2t=0\implies2t=n\pi\implies t=\dfrac{n\pi}2$

where $n$ is any integer.

$z(t)$ is increasing wherever $\sin2t>0$ , which happens for

$\dfrac{2n\pi}2$

$\boxed{n\pi$

5 0

3 years ago

составь верные равенства и неравенства, используя таблицу: 32×2, 120+253, 900-653, 300×2, 640÷8, 900÷3, 450-350, 45×3

tekilochka [14]

What is your question?

5 0

3 years ago

Find the minimum maximum and domain and range of the functions <br>h(x) = -x² + 4x - 2

Svet_ta [14]

Domain (-infinity,infinity)

Min(?

max(2,2)

3 0

3 years ago

if the equation x^2 +(k+2)x+2k=0 has equal roots,then the value of k is .....a.2 b.-2 c 1/2 d.none

blsea [12.9K]

Answer:

k=2

Problem:

if the equation x^2 +(k+2)x+2k=0 has equal roots,then the value of k is ..

Step-by-step explanation:

Since the coefficient of x^2 is 1, we can use this identity to aid us: x^2+bx+(b/2)^2=(x+b/2)^2.

So we want the following:

[(k+2)/2]^2=2k

Apply the power on the left:

(k+2)^2/4=2k

Multiply both sides by 4:

(k+2)^2=8k

Expand left side:

k^2+4k+4=8k *I used identity (x+c)^2=x^2+2xc+c^2

Subtract 8k on both sides:

k^2-4k+4=0

Factor using the identity mentioned a couple lines above:

(k-2)^2=0

Since zero squared is zero, we want k-2=0.

Adding both sides by 2 gives k=2.

8 0

3 years ago