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mr Goodwill [35]
3 years ago
11

The table gives sales data for a stationery store in any given two weeks. It lists the probability of the number of items purcha

sed in a single transaction and the average amount spent per transaction. Items Purchased in a Single Transaction Probability Average Amount Spent 2 0.35 $12 3 0.17 $20 4 0.33 $28 5 or more 0.15 $36 Based on the data provided, transactions with items are likely to bring in the most income during the next two-week transaction period.
Mathematics
1 answer:
netineya [11]3 years ago
7 0
From the given data:

Number of items | Probability of being purchased | Average amount spent
--------------------------------------------------------------------------------------------------
            2              |                     0.35                      |            $12
            3              <span>|                     0.17                      |            $20
</span>            4              <span>|                     0.33                      |            $28
</span>       5 or more      <span>|                     0.15                      |            $36
</span>
In a sample space of 100 items being purchased, in a day:

For 2 items: 35 (12) = 420
For 3 items: 17<span> (20) = 340
</span>For 4 items: <span>33 (28) = 924
</span>For 5 or more items: 15<span> (36) = 540
</span>
If this sample trend would go on the for the next two weeks, then the transaction with 4 items would most likely bring in the most income during the said transaction period.
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Answer:

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Step-by-step explanation:

  h(t) = -16t(t -18/16) . . . . put in intercept form

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Can someone help me with this and show their work and how they found the answer?
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Answer:

2kg

Step-by-step explanation:

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H = 2/3 * R

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Given that the 3 bricks weigh 5kg in total then,

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What percent of 100.9 is 82??
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Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following p
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Step-by-step explanation:

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The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 6

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           p = probability of success which in our question is 0.3.

LET X = a random variable

So, it means X ~ Binom(n=6, p=0.3)

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Therefore, P(X < 3) = 0.7443.

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