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Novosadov [1.4K]

3 years ago

11

Help me I need please

rt%7Bx%20%7B%7D%5E%7B2%7D%20%20-%209%7D%20" id="TexFormula1" title=" \sqrt{ {x}^{2} - 4 } \times \sqrt{x {}^{2} - 9} " alt=" \sqrt{ {x}^{2} - 4 } \times \sqrt{x {}^{2} - 9} " align="absmiddle" class="latex-formula">

2 answers:

DaniilM [7]3 years ago

6 0

Step-by-step explanation:

$thank \: you$

Sauron [17]3 years ago

3 0

Step-by-step explanation:

SEE THE IMAGE FOR SOLUTION ..

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What is the sum of the first five terms of a geometric series with A1 equals 20 and R equals 1/4

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Sum = 20(1-(1/4)^5) / (1-1/4) = 26.64

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The sum of 5 and twice a number.

snow_lady [41]

Answer:

5+x2

Step-by-step explanation:

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Anna71 [15]

C. The 40th customer

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3 years ago

-(-6x - 5) = 4(5x + 3)

Lina20 [59]

Answer:

$\large\boxed{B)\ -\dfrac{1}{2}}$

Step-by-step explanation:

$-(-6x-5)=4(5x+3)\qquad\text{use the distributive property}\\\\-(-6x)-(-5)=(4)(5x)+(4)(3)\\\\6x+5=20x+12\qquad\text{subtrac 5 from both sides}\\\\6x=20x+7\qquad\text{subtract 20x from both sides}\\\\-14x=7\qquad\text{divide both sides by (-14)}\\\\x=-\dfrac{7}{14}\\\\x=-\dfrac{1}{2}$

8 0

3 years ago

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

dem82 [27]

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

= 1- 0.5134

= 0.4866

6 0

3 years ago