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chubhunter [2.5K]

2 years ago

15

When a figure is translated on a coordinate grid, what conclusion can you draw from the pre-image and image?

1 answer:

lina2011 [118]2 years ago

8 0

The position of the figure changes because you are translating it. Meaning that you are moving it.

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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

Which statment is true

Paraphin [41]

I think hamburger and French fries

6 0

2 years ago

ABCD is a parallelogram and this figure is NOT drawn to scale.

dimulka [17.4K]

The angle does not matter. Think of it as finding the other side to a triangle. Use $a ^{2} +b^2 =c^2$

a=39 (line AB)
b=b (the leg we need to find)
c=89 (line BD)

$39^2 + b^2 = 89^2$
$1521 + b^2 = 7921$
(subtract the 1521 from both sides)
$b^2 = 6400$
(square root both sides)
$\sqrt{(b^2)} = \sqrt{6400} 
$

b = 80
AD=80

7 0

2 years ago

Read 2 more answers

Give the domain and range. Tell if it is a function.

Anna [14]

It’s not a function
That’s all I know sorry!

6 0

2 years ago

Point B has coordinates (4,1). The x-coordinate of point A is -4. The distance between point A and point B is 10 units. What a

klasskru [66]

Given:

Point B has coordinates (4,1).

The x-coordinate of point A is -4.

The distance between point A and point B is 10 units.

To find:

The possible coordinates of point A.

Solution:

Let the y-coordinate of point A be y. Then the two points are A(-4,y) and B(4,1).

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between point A and point B is 10 units.

$\sqrt{(4-(-4))^2+(1-y)^2}=10$

Taking square on both sides, we get

$(8)^2+(1-y)^2=100$

$(1-y)^2=100-64$

$(1-y)^2=36$

Taking square root on both sides, we get

$(1-y)=\pm \sqrt{36}$

$-y=\pm 6-1$

$y=1\mp 6$

$y=1-6$ and $y=1+6$

$y=-5$ and $y=7$

Therefore, the possible coordinates of point A are either (-4,-5) or (-4,7).

7 0

2 years ago