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Komok [63]
2 years ago
12

A group of friends were working on a student film that had a budget of $1300. They used $546 of their budget on props. What perc

entage of their total budget did they spend on props?
Mathematics
1 answer:
Katarina [22]2 years ago
5 0
To find percentage: divide the given value by the total and multiply by 100

546/1300 x 100 = 42%
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pashok25 [27]
M(CE) = 78 + 78*4/26 = 90
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3 years ago
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What is 4 times 2. Will give brainliest HELPPPPPPPPPPPPPPPPPPPPPPPp
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the answer is 8

Step-by-step explanation:

4 + 4=8

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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

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2 years ago
Make x the subject of the formula x/a -b=c
horrorfan [7]

Answer:

x ÷ a - b = c

x a-b both side

x = c(a-b)

or

x = ca - bc

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\bf \textit{difference of cubes}
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a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad
(a-b)(a^2+ab+b^2)= a^3-b^3\\\\
-------------------------------\\\\
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\\\\\\
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