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lys-0071 [83]
3 years ago
15

A company manufactured a 16-oz box of cereal. Boxes are randomly weighted to ensure the correct amount. If the discrepancy in we

ight is more than 0.15 oz, the production is stopped. What is the range of acceptable values for production to continue?
Mathematics
2 answers:
AlexFokin [52]3 years ago
4 0

Answer:

The acceptable range will be : 15.85 oz to 16.15 oz

Step-by-step explanation:

Let the weight of the box be = x

As given,

If the discrepancy in weight is more than 0.15 oz, the production is stopped.

And if the discrepancy in weight is less than 0.15 oz then the production will continue.

We can express this as : If |x-16| \leq 0.15 ; the production continues

|z| \leq a or we can say

-a\leq z\leq a

Now, z = x-16, and a=0.15, then the value becomes,

-0.15 \leq x-16 \leq 0.15

Solving for x:

We get the final range as :

-0.15+16 \leq x-16+16 \leq 0.15+16

15.85\leq x \leq 16.15

Therefore, the answer is : 15.85\leq x \leq 16.15

Katena32 [7]3 years ago
3 0
15.85 oz to 16.15 since .15 ounces less than 16 is 15.85 and .15 ounces more than 16.15
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liubo4ka [24]

The sum of  ( \frac{2}{5} x + \frac{5}{8} ) + ( \frac{1}{5} x - \frac{1}{4} ) is  \frac{3}{5} x +  \frac{3}{8}

Step-by-step explanation:

To add or subtract 2 fractions, they must have same denominators, if they not do that

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∵ ( \frac{2}{5} x + \frac{5}{8} ) + ( \frac{1}{5} x - \frac{1}{4} )

- Let us start with adding the like terms

∴  ( \frac{2}{5} x + \frac{5}{8} ) + ( \frac{1}{5} x - \frac{1}{4} ) = (

∵ ( \frac{2}{5} x + \frac{1}{5} x ) have same denominators, then we can add them

∴ ( \frac{2}{5} x + \frac{1}{5} x ) = \frac{3}{5} x

∵  ( \frac{5}{8}  - \frac{1}{4} ) do not have the same denominators, then we must find

    LCM of 8 and 4

- The LCM of 8 and 4 is 8 because 8 is the first common multiple

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∵ LCM of 8 and 4 is 8

- Divide 8 by the denominator 4

∵ 8 ÷ 4 = 2

- Multiply the numerator of the fraction by 2

∴ \frac{1}{4}=\frac{2}{8}

∴  ( \frac{5}{8}  - \frac{1}{4} ) = ( \frac{5}{8} - \frac{2}{8} ) = \frac{3}{8}

∴  ( \frac{2}{5} x + \frac{5}{8} ) + ( \frac{1}{5} x - \frac{1}{4} ) = \frac{3}{5} x +  \frac{3}{8}

The sum of  ( \frac{2}{5} x + \frac{5}{8} ) + ( \frac{1}{5} x - \frac{1}{4} ) is  \frac{3}{5} x +  \frac{3}{8}

Learn more:

You can learn more about the fractions in brainly.com/question/2456302

#LearnwithBrainly

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A factory received a shipment of 10 generators, and the vendor who sold the items knows there are 4 generators in the shipment t
jek_recluse [69]

Answer:

A) 0.026

B) 0.130

Step-by-step explanation:

Complete Question

A factory received a shipment of 10 generators and the vendor who sold the items knows there are 4 generators in the shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment, and if too many of the generators in the sample are defective, he will refuse the shipment. Give answer as a decimal to three decimal places.

(A) If a sample of 4 generators is selected, find the probability that all in the sample are defective.

(B) If a sample of 4 generators is selected, find the probability that none in the sample is defective.

Solution

A) This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of generators to be picked = 4

x = Number of successes required = number of defective generators required = 4

p = probability of success = probability that a randomly selected generator is defective = (4/10) = 0.40

q = probability of failure = probability that a randomly selected generator is NOT defective = 1 - p = 1 - 0.40 = 0.60

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B) n = total number of sample spaces = number of generators to be picked = 4

x = Number of successes required = number of defective generators required = 0

p = probability of success = probability that a randomly selected generator is defective = (4/10) = 0.40

q = probability of failure = probability that a randomly selected generator is NOT defective = 1 - p = 1 - 0.40 = 0.60

P(X = 0) = ⁴C₀ (0.40)⁰ (0.6)⁴⁻⁰ = 0.1296 = 0.130 to 3 d.p.

Hope this Helps!!!

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