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Nookie1986 [14]
3 years ago
5

Miss Taylor graded 50 English research papers. Twelve of them received a grade of A. Miss Taylor drew a circle graph to display

the grades. What percent of the circle graph was devoted to the A papers?
Mathematics
1 answer:
nikdorinn [45]3 years ago
3 0
<h3>Answer:</h3>

\large\boxed{24\%}

<h3>Step-by-step explanation:</h3>

In this question, it's asking you to find how much percentage the circle graph is for "A" papers.

To solve this question, we would need to use information from the question.

Important information:

  • Graded 50 English research papers
  • 12 of those papers had an "A" grade

With the information above, we can solve the question.

We know that there are 12 research papers that received an A and there are 50 research papers in total.

We would divide 12 by 50 in order to find the percentage of the papers that got an A.

12\div50=24

When you divide, you should get 24.

This means that 24% of the circle graph is devoted to "A" papers.

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>
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In a particular game, a fair die is tossed. If the number of spots showing is either four or five, you win $1. If the number of
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Answer:

The probability that you win at least $1 both times is 0.25 = 25%.

Step-by-step explanation:

For each game, there are only two possible outcomes. Either you win at least $1, or you do not. Games are independent. This means that the binomial probability distribution is used 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.

Probability of winning at least $1 on a single game:

The die has 6 sides.

If it lands on 4, 5 or 6(either of the three sides), you win at least $1. So

p = \frac{1}{2} = 0.5

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This means that n = 2

The probability that you win at least $1 both times is

This is P(X = 2).

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

P(X = 2) = C_{2,2}.(0.5)^{2}.(0.5)^{2} = 0.25

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Ayúdenme por favor necesito esto resolvió para mañana
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Answer:

Cómo puedo ayudar

Step-by-step explanation:

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