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mr_godi [17]
3 years ago
14

At a pizza shop, 70% of the customers order a pizza, 25% of the customers order a salad, and 15% of the customers order both a p

izza and a salad. If a customer is chosen at random, what is the probability that he or she orders either a pizza or a salad?
Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
6 0

Answer:

The probability that a randomly selected customer orders a pizza or a salad is 80%

Step-by-step explanation:

Let's call

P_p = the proability of someone asking for a pizza

P_s = the probability that a customer orders a salad

Pp∩s = the probability that a customer orders a pizza and a salad.

We know from the statement of the problem that:

P_p = 70% = 0.7

P_s = 25% = 0.25

Pp∩s = 15% = 0.15

We want to know the probability that a randomly selected customer will order a pizza or a salad. Then, by set theory, that probability is calculated as:

Pp∪s = P_p ∪ P_s - Pp∩s

So:

Pp∪s = Pp + Ps - Pp∩s

 Where Pp∪e is the probability that the customer orders a pizza or a salad, but not both at the same time.

Pp∪e = 0.7 + 0.25 -0.15

Pp∪e = 0.8

Finally, the probability that a randomly selected customer orders a pizza or a salad is 80%

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In a random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand. C
horrorfan [7]

Answer:

<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>

(0.5603,0.6529)

Step-by-step explanation:

<em>Step(i)</em>:-

<em>Given sample size 'n' =300</em>

Given data  random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand.

<em>Given sample proportion </em>

<em>                        </em>p^{-}  = \frac{x}{n} = \frac{182}{300} =0.606

level of significance = 90% or 0.10

Z₀.₁₀ = 1.645

<em>90% confidence interval for the proportion is determined by</em>

(p^{-} - Z_{0.10}\sqrt{\frac{p(1-p)}{n} }  , p^{-} +Z_{0.10}\sqrt{\frac{p(1-p)}{n} })

(0.6066 - 1.645\sqrt{\frac{0.6066(1-0.6066)}{300} }  ,0.6066+1.645\sqrt{\frac{0.6066(1-0.6066)}{300} })

(0.6066 -  0.0463  ,0.6066 +  0.0463)

(0.5603,0.6529)

<u>final answer</u>:-

<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>

(0.5603,0.6529)

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Simplify the expression.<br><br> 2+22÷4+3
Dmitrij [34]

Answer:

The answer is 10.5

Step-by-step explanation:

6 0
3 years ago
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