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yuradex [85]
3 years ago
15

A furniture store received an order for 8,367 tables they can fit 7 tables in the lower shipping box how many shipping boxes whe

re they need to ship all the tables
Mathematics
1 answer:
-BARSIC- [3]3 years ago
8 0

1195 boxes are needed to ship all the tables

<em><u>Solution:</u></em>

Given that, furniture store received an order for 8,367 tables

They can fit 7 tables in the lower shipping box

To find: Number of shipping boxes needed to ship all the tables

From given,

Total number of tables = 8367

Number of tables fit in one shipping box = 7

Thus, we can find the number of shipping boxes needed to ship all the tables by dividing total number of tables by number of tables fit in one shipping box

\text{Number of shipping boxes needed } = \frac{\text{Total number of tables}}{\text{Number of table fit in one box}}

Substituting the values we get,

\text{Number of shipping boxes needed } = \frac{8367}{7}=1195.28 \approx 1195

Thus 1195 boxes are needed to ship all the tables

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All the numbers from Bag B were written as thirds cause their probabilities could be simplified to make multiplying easier
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The Orioles played 90 games and have won 70% of them. How many games has the team won?
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Suppose a random variable, x, arises from a binomial experiment. If n = 6, and p = 0.30, find the following probabilities using
AlladinOne [14]

Answer:

<u></u>

  • <u>a) P(X=1) = 0.302526</u>
  • <u>b) P(X=5) = 0.010206</u>
  • <u>c) P(X=3) = 0.18522</u>
  • <u>d) P(X≤3) = 0.92953</u>
  • <u>e) P(X≥5) = 0.010935</u>
  • <u>f) P(X≤4) = 0.989065</u>

Explanation:

Binomial experiments are modeled by the formula:

       P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{(n-x)}

Where

  • P(X=x) is the probability of exactly x successes

  • C(n,x)=\dfrac{n!}{x!\cdot (n-x)!}

  • p is the probability of one success, which must be the same for every trial, and every trial must be independent of other trial.
  • n is the number of trials
  • 1 - p is the probability of fail
  • there are only two possible outcomes for each trial: success or fail.

<u>a.) P (x=1)</u>

<u></u>

       P(X=1)=\dfrac{6!}{1!\cdot (6-1)!}\times (0.3)^1\times(1-0.3)^{(6-1)}=0.302526

<u>b.) P (x=5)</u>

     P(X=5)=\dfrac{5!}{5!\cdot (6-5)!}\times (0.3)^5\times (1-0.3)^{(6-5)}=0.010206

<u>c.) P (x=3)</u>

Using the same formula:

    P(X=3)=0.18522

<u>d.) P (x less than or equal to 3)</u>

  • P(X≤3)= P(X=3) + P(X=2) + P(X=1) + P(X=0)

Also,

  • P(X≤3) = 1 - P(X≥4) = 1 - P(X=4) - P(X=5) - P(X=6)

You can use either of those approaches. The result is the same.

Using the second one:

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  • P(X≤3) = 1 - 0.05935 - 0.010206 - 0.000729 = 0.92953

<u>e.) P(x greather than or equal to 5)</u>

  • P(X≥5) = P(X=5) + P(X=6)
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<u>f.) P(x less than or equal 4)</u>

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