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2 years ago

9

If k people are seated in a random manner in a circle containingnchairs(n > k),whatistheprobabilitythatthe people will occupy

k adjacent chairs in the circle?

1 answer:

velikii [3]2 years ago

6 0

It depends on the amount of n and k to determine

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PPPLLLZZZ HELP I'LL GIVE BRAINLIEST

kati45 [8]

Allen's work is not written properly so I have rearranged it as shown below:

Original problem) –8.3 + 9.2 – 4.4 + 3.7.

Step 1) −8.3 + 9.2 + 4.4 + 3.7 Additive inverse

Step 2) −8.3 + 4.4 + 9.2 + 3.7 Commutative property

Step 3) −8.3 + (4.4 + 9.2 + 3.7) Associative property

Step 4) −8.3 + 17.3

We can see that in step 1), Allen changed -4.4 into +4.4 using additive inverse. Notice that we are simplifying not eliminating -4.4 as we do in solving some equation. Hence using additive inverse is the wrong step.

Alen should have collect negative numbers together and positive numbers together.

Add the respective numbers then proceed to get the answer.

–8.3 + 9.2 – 4.4 + 3.7

= –8.3 – 4.4 + 9.2 + 3.7

= -12.7 + 12.9

= 0.2

6 0

3 years ago

Read 2 more answers

A project manager has determined that of the work for a project was completed in 9 days. How many more days will it take to fini

AysviL [449]

The Correct Answer Is 6

3 0

2 years ago

What’s the 5th term is 2,14,98

iragen [17]

$\bold{\huge{\orange{\underline{ Solution }}}}$

<h3><u>Correct </u><u>Question </u><u>:</u><u>-</u></h3>

What is the 5th term of an AP 2 , 14 ....98 .

<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>

<u>We </u><u>have </u><u> </u><u>AP</u><u>, </u>

$\sf{ 2 , 14 ,.... 98 }$
<u>AP </u><u>is </u><u>the </u><u>arithmetic </u><u>progression </u><u>or </u><u>a </u><u>sequence </u><u>of </u><u>numbers </u><u>in </u><u>which </u><u>succeeding </u><u>number </u><u>is </u><u>differ </u><u>from </u><u>preceeding </u><u>number </u><u>by </u><u>a </u><u>common </u><u>value</u><u>. </u>

<h3><u>Solution </u><u>:</u><u>-</u></h3>

<u>We </u><u>have </u><u>an </u><u>AP </u><u>:</u><u>-</u><u> </u><u>2</u><u> </u><u>,</u><u> </u><u>1</u><u>4</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>9</u><u>8</u>

<u>Therefore</u><u>, </u>

$\sf{a1\: or \:1st\: term = 2 }$
$\sf{a2 \:or \:2nd\: term = 14 }$
$\sf{an \:or\: last\: term = 98}$

<u>Here</u><u>, </u>

Common difference of an AP

$\sf{ a2 - a1 }$

$\sf{ = 14 - 2}$

$\sf{ = 12}$

Thus, The common difference is 12

<u>Now</u><u>, </u>

We know that,

$\sf{an = a1 + ( n - 1 )d}$

$\sf{a5 = 2 + ( 5 - 1 )12}$

$\sf{a5 = 2 + 4 × 12}$

$\sf{a5 = 2 + 48}$

$\sf{\red{a5 = 50}}$

Hence, The 5th term of given AP is 50

4 0

2 years ago

Can someone help me this this sheet please and thank you.!

AveGali [126]

Answer:

1- faster 2- rises,sink 3-equater 4-convection idk the rest

Step-by-step explanation:

3 0

2 years ago

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Joe wants to calculate 12×8 in a calculater but the 8 key isnt working.Explain what can he do

eduard

Answer:

12x4x2

Step-by-step explanation:

4x2 =8

so 12x4x2=96

7 0

3 years ago

Read 2 more answers