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Delvig [45]
4 years ago
10

18. The price of 1 box of popcorn and 1 drink together is The price of popcorn and 1 drink together is What is the cost of 1 dri

nk on 3 q

Mathematics
2 answers:
valkas [14]4 years ago
8 0
A box of popcorn = 8.35-5.10
                            = 3.25

a drink= 5.10-3.25
           =1.85
xz_007 [3.2K]4 years ago
7 0
We know that one box of popcorn and one soda cost $5.10. We also know that two boxes of popcorn and one soda cost $8.35. To find how much a box of popcorn costs we can simply subtract $5.10 from $ 8.35. This will give us $3.25. Now we can subtract $3.25 from $5.10. This will give us $1.85. This means that a soda costs $1.85.
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7 Points: please help. What is the surface area?
faust18 [17]

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192 cm squared

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Take the following list of functions and arrange them in ascending order of growth rate. That is, if function g(n) immediately f
zepelin [54]

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4 0
3 years ago
Suppose it is known that 60% of radio listeners at a particular college are smokers. A sample of 500 students from the college i
vladimir1956 [14]

Answer:

The probability that at least 280 of these students are smokers is 0.9664.

Step-by-step explanation:

Let the random variable <em>X</em> be defined as the number of students at a particular college who are smokers

The random variable <em>X</em> follows a Binomial distribution with parameters n = 500 and p = 0.60.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=500\times 0.60=300>10\\n(1-p)=500\times(1-0.60)=200>10

Thus, a Normal approximation to binomial can be applied.

So,  

X\sim N(\mu=600, \sigma=\sqrt{120})

Compute the probability that at least 280 of these students are smokers as follows:

Apply continuity correction:

P (X ≥ 280) = P (X > 280 + 0.50)

                   = P (X > 280.50)

                   =P(\frac{X-\mu}{\sigma}>\frac{280-300}{\sqrt{120}}\\=P(Z>-1.83)\\=P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that at least 280 of these students are smokers is 0.9664.

8 0
3 years ago
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