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

horrorfan [7]

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 )

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