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katovenus [111]
2 years ago
8

A bottle of detergent that will wash 32 loads of laundry will cost 4.80. A bottle that will wash 50 loads cost 7.00.which is the

better deal? prove your answer
Mathematics
1 answer:
max2010maxim [7]2 years ago
5 0

32. Because if you do 4.80+4.80 it’s 9.60which is less than 7.00+7.00

Step-by-step explanation:

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Step-by-step explanation:

yf(x)^{-1}  = inverse\\f(x)=y \\y = 1/(x^{3} \\Inverse: y=x ------------> x = 1/y^{3}\\y^{3} - \frac{1}{x} = 0\\y^{3} = \frac{1}{x}\\y = \sqrt[3]{\frac{1}{x}} \\y = \frac{\sqrt[3]{1} }{\sqrt[3]{x}} \\y = \frac{1}{\sqrt[3]{x}}

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The area of the entire figure below is 1 square unit.
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Refer to the Trowbridge Manufacturing example in Problem 2-35. The quality control inspection proce- dure is to select 6 items,
Ivanshal [37]

Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The true proportion of defects is 0.15

This means that p = 0.15

Sample of 6:

This means that n = 6

What is the probability that there will be 0 or 1 defects in a sample of 6?

P(X \leq 1) = P(X = 0) + P(X = 1)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{6,0}.(0.15)^{0}.(0.85)^{6} = 0.3771

P(X = 1) = C_{6,1}.(0.15)^{1}.(0.85)^{5} = 0.3993

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764

77.64% probability that there will be 0 or 1 defects in a sample of 6.

5 0
3 years ago
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