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KATRIN_1 [288]
2 years ago
7

This problem illustrates the limit derivation of a Poisson distribution from Binomial distributions. Suppose an average of 6 arr

ivals occur during a 30 minute interval. To count arrivals, divide the 30 minute interval into n sub-intervals. Enter the probability p of one arrival during a single sub-interval.
For n = 30, we have p =
For n = 60, we have p =
For n = 100, we have p =
Continued from previous.
Mathematics
1 answer:
expeople1 [14]2 years ago
7 0

Answer:

I need help with mathematics please

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Plzzz help me with #3
solmaris [256]

Answer:

The answer to question 3 is B.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Quadrilateral ABCD has vertices A (-3, 0), B (2, 4), C (3, 1), and D (-4, -3). Calculate the perimeter of the quadrilateral. Rou
Andru [333]

Answer:

Perimeter of the Quadrilateral=20.78

Step-by-step explanation:

Perimeter of the Quadrilateral= AB+BC+CD+AD

Finding the sides of the quadrilateral using the Distance formula:

A(-3,0), B(2,4), C(3,1) and D(-4,-3)

AB=\sqrt{(2-(-3))^2+(4-0)^2}\\\\ =\sqrt{(2+3)^2+4^2}\\\\ =\sqrt{5^2+4^2} \\\\=\sqrt{25+16}\\\\ =\sqrt{41}\\

=6.40

BC=

         \sqrt{(3-2)^2+(1-4)^2} \\\\=\sqrt{1^2+(-3)^2} \\\\=\sqrt{1+9}\\\\ =\sqrt{10}

          =3.16

      CD=\sqrt{(-4-3)^2+(-3-1)^2} \\\\=\sqrt{(-7)^2+(-4)^2} \\\\=\sqrt{49+16}\\\\=\sqrt{65}\\\\ =8.06

AD=\sqrt{(-4-(-3))^2+(-3-0)^2} \\\\=\sqrt{(-4+3)^2+(-3)^2} \\\\=\sqrt{(-1)^2+(-3)^2} \\\\=\sqrt{1+9}\\\\=\sqrt{10} \\\\ =3.16

Perimeter of the quadrilateral= AB+BC+CD+AD

             =6.40+3.16+8.06+3.16

              =20.78

6 0
3 years ago
How do you solve this?
dalvyx [7]
P=2(L+W)
L=2/3x+10
W=1/3x+5
P=2(2/3x+10+1/3x+5)
P=2(2/3x+1/3x+10+5)
P=2(3/3x+15)
P=2(x+15)
P=2x+30
6 0
3 years ago
Benito runs 1/10 of a mile each day. Which shows how to find the number of days it will take for Benito to run 3/5 of a mile?
n200080 [17]

D is most likely right. Basically, I would turn them into decimals and divide using a graphing calculator.

4 0
3 years ago
If an object is projected upward with an initial velocity of 123 ft per​ sec, its height h after t seconds is h=−16t2+123t. Find
Alinara [238K]

Answer:

Height after 5 seconds is <em>215 ft</em>

Step-by-step explanation:

Given that:

The initial velocity of object which is projected upwards is 123 ft/sec.

Height of the object, <em>h</em> after time <em>t </em> in seconds,  is given as:

h=-16t^2+123t

Here, <em>t </em>will always be positive, so 123t and 16t^2 will also be positive.

But coefficient of 16t^2 is negative, that means something is subtracted from the positive term 123t.

To find:

Height of object after 5 seconds.

Solution:

Given that t=5 seconds.

Let us put the value in the given relation of h and t:

h=-16\times 5^2+123 \times 5\\\Rightarrow h=-400+615\\\Rightarrow \bold{h = 215\ ft}

So, height after 5 seconds is <em>215 ft</em>.

6 0
3 years ago
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