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nlexa [21]
3 years ago
6

4 friends evenly divided up an n-slice pizza. One of the friends, Harris, ate 1 fewer slice than he received. How many slices of

pizza did Harris eat? Write your answer as an expression. Pls answer fast I'm taking a test!!
Mathematics
1 answer:
atroni [7]3 years ago
6 0

Answer:

7

Step-by-step explanation:

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Jason helped his dad put plastic and aluminum items into the recycling container. The container is a large cube with a length, w
igor_vitrenko [27]

Answer:

its 216 cubic feet.

Step-by-step explanation:

6x6x6 is 216

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Identify the terms themselves for the following expression. (separate ur terms with a comma)
Zinaida [17]
3a, 5b, -6y, 10 hope this helps
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3 years ago
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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
3 years ago
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Helen [10]
Table C is a function, because none of the numbers in the X column repeat. (All the other ones do.):-)Hope I could help!!
4 0
3 years ago
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PLEASE HELP!! ASAP!!
mr Goodwill [35]

Answer:

D and E

Step-by-step explanation:

A vertical asymptote in this case would mean that the denominator is equal to zero, which is undefined. You can factor the denominator to get you the fraction:

x+2 / (x+6)(x-4) = 0

Then you know that to equal zero, x would have to equal either -6 or 4, which is E and D.

6 0
3 years ago
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