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2 years ago

12

An elf, a dwarf, a hobbit, a wizard, and a man are sitting in a line.How many different ways can they sit if the elf and the dwa

rf insist on sitting next to each other

1 answer:

Oksana_A [137]2 years ago

6 0

Answer: 24

Step-by-step explanation:

1. Number Them

Elf and Dwarf = 1

Hobbit = 2

Wizard = 3

Man = 4

* Elf and Dwarf are put together so that they are always sitting together

2. Multiply

1 x 2 x 3 x 4 = 24

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You have to turn 65 into a decimal. That would be .65
They do 39/.65
You should get 60 as your final answer.

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HELP PLEASE!!! Thanks!

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Answer:

2) $\frac{\sqrt{77}}{11}$

3) 5√2

Step-by-step explanation:

Simplest radical form of an expression is the expression in radical so that there are no more square roots, cube roots, 4th roots, etc left to find, after rationalising the denominator ( if needed ).

2. Here the given expressions,

$\frac{\sqrt{7}}{\sqrt{11}}$

For rationalising the denominator multiply both numerator and denominator by √11,

$\frac{\sqrt{77}}{11}$

3. given expression,

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$=\sqrt{25\times 2}$

$=\sqrt{25}\times \sqrt{2}$ $(\because \sqrt{ab}=\sqrt{a}\sqrt{b})$

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3 years ago

If f(x) = 2x2^2 + 1 and g(x) = x^2 - 7, find (f+g) (x)

Sveta_85 [38]

(f+g)(x) = f(x) + g(x)

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3 years ago

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

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Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

c)

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

d)

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

<u>For Bus:</u>

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

<u>For motorcycle:</u>

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

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