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serg [7]
3 years ago
11

Someone help with #1

Mathematics
2 answers:
Drupady [299]3 years ago
6 0
My answer was wrong, please delete this
snow_lady [41]3 years ago
4 0
Ok, first we set up 2 equations with the given info
let chips = c & pretzels = p

5c + 9p = 16.95
10c + 10p = 22.50

next we solve for a single variable in one of the equations (I chose the first one)
c =  \frac{16.95 - 9p}{5}

now plug it into our second equation
33.90 - 18p + 10p = 22.50
simplify
11.40 = 8p
divide
1.425 = p

next plug p back into our first equation
5c + 9(1.425) = 16.95
5c = 4.125
c = .825

Chips cost = $0.825
Pretzels cost = $1.425



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Question Help When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select
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Answer:

The probability is P(X  \le 2 ) = 0.9072

The company will accept 90.72% of the shipment and will reject (100 -90.72) =  9.2\% of the shipment , so many of the shipment are rejected

Step-by-step explanation:

From the question we are told that

   The  sample size is  n =  36

   The  proportion that did not meet the requirement is  p =  0.03

Generally the probability that the whole shipment is accepted is equivalent to the probability that there is at most 2 batteries that do not meet the requirement , this is mathematically represented as

     P(X  \le 2 ) = [ P(X =  0 ) +  P(X =  1 ) + P(X = 0)]

=>   P(X  \le 2 ) = [ [^{n}C_0 * (p)^{0} *(1-p)^{n-0} ] +  [^{n}C_1 * (p)^{1} *(1-p)^{n-1} ] +  [^{n}C_2 * (p)^{2} *(1-p)^{n-2} ]]

Here C stands for Combination (so we will be making the combination function in our  calculators )

So

=> P(X  \le 2 ) = [ [^{36}C_0 * (0.03)^{0} *(1-0.03)^{36-0} ] +  [^{36}C_1 * (0.03)^{1} *(1-0.03)^{36-1} ] +  [^{36}C_2 * (0.03)^{2} *(1-0.03)^{36-2} ]]

=> P(X  \le 2 ) = [ [1 * 1 * 0.3340  ] +  [36* 0.03 *0.3444 ] +  [630 * 0.0009 *(0.355 ]]

=>P(X  \le 2 ) = 0.9072

The company will accept 90.72% of the shipment and will reject (100 -90.72) =  9.2\% of the shipment , so many of the shipment are rejected

 

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