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77julia77 [94]
3 years ago
9

(3 pt) Marie has $830 in her bank account and withdraws $60 each month. Denise has $970 in her bank account and withdraws $80 ea

ch month. In how many months will Marie and Denise have equal amounts of money in their accounts? A. 4 B. 5 C. 7 D. 9
Mathematics
2 answers:
tankabanditka [31]3 years ago
8 0

C. 7 is your answer  



Hope this helps & good luck. :)


Lady bird [3.3K]3 years ago
3 0
The answer is c because if you do 7x60 you'll get 420 and 830-420=410 and 7x80 you'll get 560 and 970-560 is also 410.
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Answer:

If you want me to solve for b then

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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

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         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,

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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)

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                                \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}

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So,

P(No collision in 6 months period)

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so,

P(1 or fewer collision in 6 months period)

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7 0
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Answer:

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Step-by-step explanation:

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