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3 years ago

7

How many 3-digit numbers contain exactly one 1?

2 answers:

Tems11 [23]3 years ago

8 0

Given a 3-digit number containing exactly one "1".
(ASSUME base-10 numbers, so digits are from 0-9)

Number contains exactly one "1", therefore the two other digits are limited to 2,3,4,....9,0.
If the "1" is in the leftmost position, there are 9*9=81 ways to arrange the second and third positions.
If the "1" is in the second or third position, then there are 8 choices for the leftmost digit (2-9), and 9 choices for the other digit for a total of 2*8*9=144 arrangements.
So the total number of 3-digit numbers containing exactly one "1" is 81+144=225.

marysya [2.9K]3 years ago

6 0

I believe the answer would be 17, because 100, 210, 201, 301, 310, 401, 410, 501, 510, 601, 610, 701, 710, 801, 810, 901, 910

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8n^2-4(7m^3+5n^2)<br> help

Rasek [7]

Hello there!

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3 years ago

An urn contains 12 balls, of which 4 are white.

wel

Answer:

a) $\frac{1}{3}$

b) $\frac{4}{12}\,,\,\frac{4}{11}\,,\,\frac{4}{10}\,,...$

Step-by-step explanation:

Number of balls in the urn = 12

Number of white balls in the urn = 4

So, number of balls which are not white = 12 - 4 = 8

We know that probability = No. of outcomes/Total number of outcomes

Let $A_i\,,\,B_i\,,\,C_i$ denotes the event of getting white ball by A, B and C

a) each ball is replaced after being drawn.

Probability = Number of white balls/Total number of balls

Solution: $P(A_i)=P(B_i)=P(C_i)=\frac{4}{12}=\frac{1}{3}$

b)the balls that are withdrawn are not replaced.

Solution:

If A wins: $P(A_1)=\frac{4}{12}=\frac{1}{3}$

If A lost and B wins: $P\left ( B_1|A_1^{c} \right )=\frac{4}{11}$

If A and B lost and C wins: $P\left ( C_1|A_1^{c}\,B_1^{c} \right )=\frac{4}{10}$

and so on....

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3 years ago

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Alex17521 [72]

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Step-by-step explanation:

i got it right when i looked it up lol

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3 years ago

With full explanation from the internet like before<br> 3x2-6x+5=0

Elodia [21]

Answer:

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Explanation:

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3x² - 6x + 5 = 0

Use the Quadratic Formula:

$\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \ \ when \ \ ax^2 + bx + c = 0$

<u>insert coefficients</u>

$\Longrightarrow \sf x = \dfrac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}$

$\Longrightarrow \sf x = \dfrac{\left6\right\pm \sqrt{-24} }{6}$

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2 years ago

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Zinaida [17]

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4 years ago