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2 years ago

Okay can you please help me do this

Mathematics

1 answer:

sertanlavr [38]2 years ago

7 0

Answer:

y=3

Step-by-step explanation:

Multiply 9 from both sides to cancel it out.

-27=-y/9

/9 /9

-3=-y

We can't have a negative variable, so we divide -1 from both sides.

-3=-y

/-1 /-1\

3=y

-hope it helps

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If g(x) = x+1/x-2 and h(x) = 4-x, what is the value of (g•h)(-3)?

Yakvenalex [24]

Answer:

I think my answer is

C. 15/2

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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

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

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

P(zero exact change in 10 minutes) = 0.4347

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

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>

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

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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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3 years ago

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spin [16.1K]

Answer:

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Step-by-step explanation:

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2 years ago

3 times the measure of an angle is 14 less than the measure of its complement. What is the measure of the angle?

Sedbober [7]

Let the measure of the angle be x and from that the measure of its complement is 90 - x. The equation that best represent the given condition above is,
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3 years ago

Write an equation in standard form for a line that passes through (4,1) & (5,7)

tatuchka [14]

Y = mx + b

First equation 1 = m(4) + b, bring everything to one side m(4) + b - 1 = 0
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Set them equal to each other,

m(4) + b - 1 = m(5) + b - 7

If you bring the b over to the left hand side it becomes

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6 = m

Plug m = 6 into either equation from the beginning,

m(4) + b - 1 = 0

6(4) + b - 1 = 0

24 + b - 1 = 0

b = -23

Knowing m and b we can now make an equation

y = mx + b

y = 6x -23 Final answer

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3 years ago