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2 years ago

Static and Reasoning:

Mathematics

1 answer:

Serggg [28]2 years ago

5 0

Based on the data recorded by Isabella, it can be concluded the cube is rather fair.

<h3>How many times did Isabella get each number?</h3>

Based on the data, here are the results:

Getting a 1: 6 times
Getting a 2: 5 times
Getting a 3: 7 times
Getting a 4: 5 times
Getting a 5: 6 times
Getting a 6: 7 times

This implies, in total Isabella got the same number between five and seven times. For example, the number 2 was obtained 5 times, but the number 3 was obtained 7 times.

<h3>What can be concluded based on the results?</h3>

Even though Isabella did not get the same number of times each number, the dice is rather fair because by rolling the dice thirty six times you will obtain the same number at least five times.

Moreover, there is not a big difference in the number of times you obtain each number.

Learn more about dice in: brainly.com/question/23637540

#SPJ1

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A company that produces fine crystal knows from experience that 17% of its goblets have cosmetic flaws and must be classified as

Alex73 [517]

Answer:

0.3891 = 38.91% probability that only one is a second

Step-by-step explanation:

For each globet, there are only two possible outcoes. Either they have cosmetic flaws, or they do not. The probability of a goblet having a cosmetic flaw is independent of other globets. 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.

17% of its goblets have cosmetic flaws and must be classified as "seconds."

This means that $p = 0.17$

Among seven randomly selected goblets, how likely is it that only one is a second

This is P(X = 1) when n = 7. So

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

$P(X = 1) = C_{7,1}.(0.17)^{1}.(0.83)^{6} = 0.3891$

0.3891 = 38.91% probability that only one is a second

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The answer is 29 I put it in my scientific calculator

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Step-by-step explanation:

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