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ANTONII [103]
1 year ago
14

the advertised weight of a snickers fun size bar is 17 grams. what proportion of candy bars in this sample weigh less than adver

tised? (round your answer to three decimal places.)
Mathematics
1 answer:
irakobra [83]1 year ago
6 0

0.412 proportion of candy bars in this sample weigh less than advertised.

<h3>What do you mean by Proportion?</h3>

The size or a number of one things (groups) as compared to that of another thing.

For example:

1. The proportion of cars and bike in your home is 4/5

2. The proportion of boy and girl in a classroom is 1/2.

3. The proportion of sea and land 3/4.

4. The proportion of kinetic energy and potential energy is 5:2.

From the given list of data:

Remove the weights which are less than the weight of the advertised weight of the snickers fun.

we will get 7g

now, the proportion will be: 7g/17g

which is 0.412

Hence,

0.412 proportion of candy bars in this sample weigh less than advertised.

Learn more about " Proportion " from here: brainly.com/question/2548537

#SPJ4

Disclaimer: The question given was incomplete on the portal. Here is the complete question.

Question: the advertised weight of a snickers fun size bar is 17 grams. what proportion of candy bars in this sample weigh less than advertised? (round your answer to three decimal places.) See the attached figure for the data.

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Answer:

The reuired probability is 0.756

Step-by-step explanation:

Let the number of trucks be 'N'

1) Trucks on interstate highway N'= 76% of N =0.76N

2) Truck on intra-state highway N''= 24% of N = 0.24N

i) Number of trucks flagged on intrastate highway  = 3.4% of N'' = \frac{3.4}{100}\times 0.24N=0.00816N

ii)  Number of trucks flagged on interstate highway  = 0.7% of N' = \frac{0.7}{100}\times 0.76N=0.00532N

Part a)

The probability that the truck is an interstate truck and is not flagged for safety is P(E)=P_{1}\times (1-P_{2})

where

P_{1} is the probability that the truck chosen is on interstate

P_{2} is the probability that the truck chosen on interstate is flagged

\therefore P(E)=0.76\times (1-0.00532)=0.756

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