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Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

2 years ago

A ball is thrown from a height of 139 feet with an initial downward velocity of 15 ft/s. The ball's height h (in feet) after t s

ella [17]

Answer:

t = 2.52 seconds

Step-by-step explanation:

h=139-15t-16t^2

We want to know when the ball hits the ground

That would be when h=0

0 = 139-15t-16t^2

We can use the quadratic formula to find t

t = -b ± sqrt(b^2-4ac)

----------------------

2a

where a = -16 b = -15 and c = 139

t = -(-15) ± sqrt((-15)^2-4(-16)139)

----------------------

2(-16)

t = (15) ± sqrt(225+8896)

----------------------

-32

t = (15) ± sqrt(9121)

----------------------

-32

t = 15+ sqrt(9121) t = 15- sqrt(9121)

-------------------- or -------------------

-32 -32

-3.453247707 or 2.515747707

Since time cannot be negative

2.515747707

Round to the nearest hundredth

t = 2.52 seconds

6 0

3 years ago

What is the value of 4 x minus 7 when x = 4?

Stolb23 [73]

Answer: 9

Step-by-step explanation: Here, we have the expression 4x - 7 and we want to evaluate the expression when x is equal to 4.

To evaluate an expression, we simply plug the value

of the variable into the expression and solve.

So here, since x is equal to 4, we have 4(4) - 7.

4(4) is equal to 16.

So we have 16 - 7 which is equal to 9.

So the value of our expression when x is equal to 4 is 9.

6 0

2 years ago

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Can y’all Plz help me ??(Show ya work) It Says What Value of X makes this equation true

olga2289 [7]

Answer:

$0.6x - 0.1x = 7 + 5 \\ 0.5x = 12 \\ x = \frac{12}{0.5} = 24$

7 0

3 years ago

Solve the quadratic equation 3x² + 2x-4=0 Give your answers to 2 decimal places.

GalinKa [24]

3x² + 2x - 4 = 0

Δ = b² - 4.a.c

Δ = 2² - 4 . 3 . -4

Δ = 4 - 4. 3 . -4

Δ = 52

x = (-b +- √Δ)/2a

x' = (-2 + √52)/2.3

x'' = (-2 - √52)/2.3

x' = 0,8685170918213297

x'' = -1,5351837584879966

2 Decimal places

x' = 0,87

x'' = -1,54

6 0

3 years ago

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