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svet-max [94.6K]
2 years ago
5

Grayson measured a line to be 10.9 inches long. If the actual length of the line is 13.4 inches, then what was the percent error

of the measurement, to the nearest tenth of a percent?
Mathematics
1 answer:
aksik [14]2 years ago
7 0

The percent error of the measurement, to the nearest tenth of a percent is 18.6%

<h3>Calculating percent error</h3>

The formula for calculating the percent error is expressed as:

%error = actual - measure/actual * 100

Given the following

actual length = 13.4 in

Measured =10.9in

Substitute

%error = 13.4-10.9/13.4 * 100
%error = 2.5/13.4 * 100
%error = 18.65%

Hence the percent error of the measurement, to the nearest tenth of a percent is 18.6%

Learn more on percent error here; brainly.com/question/1600371

#SPJ1

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After an extensive advertising campaign, the manager of a company expects the proportion of potential customers that recognize a
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Answer:

Step-by-step explanation:

Hello!

The study variable is:

X: number of customers that recognize a new product out of 120.

There are two possible recordable outcomes for this variable, the customer can either "recognize the new product" or " don't recognize the new product". The number of trials is fixed, assuming that each customer is independent of the others and the probability of success is the same for all customers, p= 0.6, then we can say this variable has a binomial distribution.

The sample proportion obtained is:

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Considering that the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample proportion to normal: p' ≈ N(p;\sqrt{\frac{p(1-p)}{n} })

The other conditions for this approximation are also met: (n*p)≥5 and (n*q)≥5

The probability of getting the calculated sample proportion, or lower is:

P(X≤0.45)= P(Z≤\frac{0.45-0.6}{\sqrt{\frac{0.6*0.4}{120} } })= P(Z≤-3.35)= 0.000

This type of problem is for the sample proportion.

I hope this helps!

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3 years ago
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Answer:

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Step-by-step explanation:

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