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GrogVix [38]
2 years ago
9

Charlie's teacher claims that he does not study you just guess it on the exam with 201 true-false questions Charlie answers 53.7

% of equations correctly calculator calculating using these results show that if we were really guessing they would be probably one in one chance and 7 that he would do well in this difficult evidence significant evidence that Charlie is just guessing why or why not​
Mathematics
1 answer:
AlladinOne [14]2 years ago
3 0

Answer: No there isn't

Explanation:

A score of 53.7% is quite close to 50% and this is a true or false exam. Charlie could have easily gotten this result by indeed guessing and not studying. This test mark is therefore not high enough to disregard the teachers's claim. Were the results to be significantly high enough above 50% then it could be said that indeed Charlie does study for his exams.

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2/3 divided by 2 1/2=????<br><br><br>in simplest form <br><br><br>ASAP PLEASE
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2 years ago
A manufacturer of processing chips knows that 2\%2%2, percent of its chips are defective in some way. Suppose an inspector rando
kipiarov [429]

The data in the question seems a bit erroneous. I am writing the correct question below:

A manufacturer of processing chips knows that 2%, percent of its chips are defective in some way. Suppose an inspector randomly selects 4 chips for an inspection. Assuming the chips are independent, what is the probability that at least one of the selected chips is defective? Lets break this problem up into smaller pieces to understand the strategy behind solving it.

Answer:

The probability that at least one of the selected chips is defective is 0.0776.

Step-by-step explanation:

The question states that the probability of defective chips is 2% i.e. 0.02. Let p denote the probability of selecting a defective chip so, p = 0.02

An inspector selects 4 chips, which means n=4 and we need to compute the probability that at least one of the selected chips is defective. Let X be the number of defective chips selected. We need to compute P(X≥1) which means either 1, 2, 3 or 4 chips can be defective.

We will use the binomial distribution formula to solve this problem. The formula is:

<u>P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ</u>

where n = total no. of trials

          p = probability of success

          x = no. of successful trials

          q = probability of failure = 1-p

we have n=4, p=0.02 and q=1-0.02=0.98.

We need to compute P(X≥1) which is equal to:

P(X≥1) = P(X=1) + P(X=2) + P(X=3) + P(X=4)

A shorter method to do this is to use the total probability theorem:

P(X≥1) = 1 - P(X<1)

          = 1 - P(X=0)

          = 1 - ⁴C₀ (0.02)⁰(0.98)⁴⁻⁰

          = 1 - (0.98)⁴

          = 1 - 0.9224

P(X≥1) = 0.0776

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3 years ago
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A number that is negative is less than 0, so

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