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sveticcg [70]
3 years ago
8

Tim bought 5 new baseball trading cards to add to his collection. The next day his dog ate half of his collection. There are now

only 27 cards left. How many cards did Tim start with?
Mathematics
1 answer:
Lunna [17]3 years ago
7 0

So if the dog ate half his collection and Tim now had 27 cards left if we times it by two we will get the number of cards he had including the 5 new ones so:

27 x 2 = 54 and then if we take the 5 new ones away we get 49 so Tim had 49 cards at the start

ANSWER: 49 cards

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Step-by-step explanation:

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Ur missing numbers are -2, 6, and 36

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4 years ago
A film distribution manager calculates that 7% of the films released are flops. If the manager is correct, what is the probabili
diamong [38]

Answer:

0.9909 = 99.09% probability that the proportion of flops in a sample of 404 released films would be greater than 4%

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.07, n = 404. So

\mu = E(X) = np = 404*0.07 = 28.28

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{404*0.07*0.93} = 5.13

If the manager is correct, what is the probability that the proportion of flops in a sample of 404 released films would be greater than 4%?

This is the pvlaue of Z when X = 0.04*404 = 16.16. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{16.16 - 28.28}{5.13}

Z = -2.36

Z = -2.36 has a pvalue of 0.0091

1 - 0.0091 = 0.9909

0.9909 = 99.09% probability that the proportion of flops in a sample of 404 released films would be greater than 4%

7 0
4 years ago
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