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katen-ka-za [31]

3 years ago

13

A lock has a 3-number code made up of 15 numbers. If none of the numbers are allowed to repeat, how many different ways can you

choose three different numbers in order for a unique code?

1 answer:

Ulleksa [173]3 years ago

3 0

Answer:2730

Step-by-step explanation:

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A random draw is being designed for 210 participants. A single winner is to be chosen, and all the participants must have an equ

Over [174]

Answer: The correct number of balls is (b) 4.

Step-by-step explanation: Given that a single winner is to be chosen in a random draw designed for 210 participants. Also, there is an equal probability of winning for each participant.

We are using 10 balls, numbered through 0 to 9. We are to find the number of balls which needs to be picked up, regardless of order, so that each of the 210 participants can be assigned a unique set of numbers.

Let 'r' represents the number of balls to be picked up.

Since we are choosing from 10 balls, so we must have

$^{10}C_r=210.$

The value of 'r' can be any one of 0, 1, 2, . . , 10.

Now,

if r = 1, then

$^{10}C_1=\dfrac{10!}{1!(10-1)!}=\dfrac{10!}{1!9!}=\dfrac{10\times 9!}{1\times 9!}=10$

If r = 2, then

$^{10}C_2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10\times 9\times 8!}{2\times 1\times 8!}=45$

If r = 3, then

$^{10}C_3=\dfrac{10!}{3!(10-3)!}=\dfrac{10!}{3!7!}=\dfrac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}=120$

If r = 4, then

$^{10}C_4=\dfrac{10!}{4!(10-4)!}=\dfrac{10!}{4!6!}=\dfrac{10\times 9\times 8\times\times 7\times 6!}{4\times 3\times 2\times 1\times 6!}=210.$

Therefore, we need to pick 4 balls so that each participant can be assigned a unique set of numbers.

Thus, (b) is the correct option.

4 0

3 years ago

Read 2 more answers

R560 invested at 8% p.a. simple interest for 3 years

alexdok [17]

Answer:

R 694.4

Explanation:

simple interest = $\sf{\dfrac{PRT}{100}$ → $\sf \dfrac{560*8*3}{100}$ → R 134.4

where p refers to principal, r refers to rate, t refers to time.

total money:

invested money + simple interest

R 134.4 + R 560 → R 694.4

4 0

2 years ago

Read 2 more answers

Can anybody help <br> 48=x^2+2x

lubasha [3.4K]

Answer:

$x^{2} + 2x - 48 = 0$

-> x = 6, x = -8

4 0

3 years ago

Bryan hiked up to the top of City Creek in 3 hr and then returned down the canyon to the trailhead in another 2 hr. His speed do

kati45 [8]

Still need your answer ?

7 0

3 years ago

Please Answer!!!!

ExtremeBDS [4]

First mug holds the most

<em><u>Solution:</u></em>

Given that,

You are choosing between two mugs

<em><u>The volume of cylinder is given as:</u></em>

$V = \pi r^2h$

Where,

r is the radius and h is the height

<em><u>One has a base that is 5.5 inches in diameter and a height of 3 inches</u></em>

$radius = \frac{diameter}{2}$

Therefore,

$r = \frac{5.5}{2}\\\\r = 2.75$

Also, h = 3 inches

<em><u>Thus volume of cylinder is given as:</u></em>

$V = \pi \times 2.75^2 \times 3\\\\V = 3.14 \times 22.6875\\\\V = 71.23875 \approx 71.24$

Thus first mug holds 71.24 cubic inches

<em><u>The other has a base of 4.5 inches in diameter and a height of 4 inches</u></em>

$Radius = \frac{4.5}{2}\\\\r = 2.25$

h = 4 inches

Therefore,

$V = 3.14 \times 2.25^2 \times 4\\\\V = 3.14 \times 20.25\\\\V = 63.585$

Thus the second mug holds 63.585 cubic inches

On comparing, volume of both mugs,

Volume of first mug > volume of second mug

First mug holds the most

8 0

3 years ago