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Drupady [299]
3 years ago
10

A basket contains 11/12 pound of strawberries. One serving strawberries is 1/4 pound. How many servings of strawberries are in t

he basket?
NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!!!
Mathematics
2 answers:
kirill115 [55]3 years ago
8 0
  • Weight of strawberries=11/12pound
  • Weight of one serving=1/4pound

☆No of servings =Total strawberries weight/Per servings weight

\\ \sf\longmapsto \dfrac{11}{12}\div \dfrac{1}{4}

\\ \sf\longmapsto \dfrac{11}{12}\times \dfrac{4}{1}

\\ \sf\longmapsto \dfrac{11(4)}{12}

\\ \sf\longmapsto \dfrac{44}{12}

\\ \sf\longmapsto \dfrac{11}{3}

  • There are total 11/3servings
egoroff_w [7]3 years ago
5 0

Answer:

3 2/3 servings

Step-by-step explanation:

Take the total amount of strawberries and divide by the number of strawberries in a serving.

11/12÷ 1/4

Copy dot flip

11/12 * 4/1

Rewriting

11/1 * 4/12

11/1 *1/3

11/3

Changing to a mixed number

3 2/3 servings

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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
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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!}


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Now, since we are calculating  this for 4 months,

so, P(No collision in 4 month period)

     =0.7408182207^{4}

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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}

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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)          

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