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Shtirlitz [24]
3 years ago
11

Serena had yards of tape. She used yards to attach a sign to the wall. How many yards of tape does Serena have left? Express you

r answer in simplest form. yards
Mathematics
1 answer:
professor190 [17]3 years ago
6 0
Serena had 7 4/9 yd of tape.
She used 2 1/9 yd.
How much tape does she have left?

        7 4/9
     -  2 1/9
--------------

Since both fractions have the same denominator, and since the lower fraction is smaller than the upper fraction, you can subtract the fractions by subtracting the numerators and using the same denominator. Then also subtract the whole numbers.

        7 4/9
     -  2 1/9
--------------
        5 3/9

The difference is 5 3/9.
3/9 can be reduced since both 3 and 9 are divisible by 3, so we divide both 3 and 9 by 3 to get 1/3.

Answer: 5 1/3 yards
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Answer:

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Step-by-step explanation:

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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
Can someone help me with this problem.
Kryger [21]
It would be 8√17 because the seventeens don't change only the one that subtracting 9.
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3 years ago
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Anit [1.1K]

Answer:

subject?

Step-by-step explanation:

8 0
2 years ago
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