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Margaret [11]
3 years ago
11

For Lilly‘s birthday she takes her for friends to the movies she buys herself all of her friends the movie tickets which cost 72

5 each at the concession stand she buys two types of popcorn for 850 and three large sodas for 575 since it's her birthday and the concession stand employer gives earlier 15% discount how much you guys understand how much does Lily spend
Mathematics
1 answer:
VladimirAG [237]3 years ago
5 0

Answer:

6536.25

Step-by-step explanation:

Given that :

Number of friends = 4

Cost of movie ticket = 725 each

At concession stand:

Pop corn = 850

Number of popcorn = 2

Large soda = 575

Number of large sodas = 3

Discount given = 15%

Amount spent on movie ticket :

725 * 5 = 3625

Amount spent at concession stand :

Cost of popcorn + cost of large soda

(850 * 2) + (575 * 3) = 3425

15% discount on purchase :

Discount = ((1-0.15) * 3425) = 2911.25

Total amount spent :

3625 + 2911.25

= 6536.25

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3 years ago
A simple random sample of size nequals=8181 is obtained from a population with mu equals 77μ=77 and sigma equals 27σ=27. ​(a) De
ivanzaharov [21]

Answer:

a) P(\bar X>81.5)=1-0.933=0.067

b) P(\bar X

c) P(73.4  

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

X \sim N(\mu=77,\sigma=27)  

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

On this case  \bar X \sim N(77,\frac{27}{\sqrt{81}})

Part a

We want this probability:

P(\bar X>81.5)=1-P(\bar X

The best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

If we apply this formula to our probability we got this:

P(\bar X >81.5)=1-P(Z

P(\bar X>81.5)=1-0.933=0.067

Part b

We want this probability:

P(\bar X\leq 69.5)

If we apply the formula for the z score to our probability we got this:

P(\bar X \leq 69.5)=P(Z\leq \frac{69.5-77}{\frac{27}{\sqrt{81}}})=P(Z

P(\bar X\leq 69.5)=0.0062

Part c

We are interested on this probability

P(73.4  

If we apply the Z score formula to our probability we got this:

P(73.4

=P(\frac{73.4-77}{\frac{27}{\sqrt{81}}}

And we can find this probability on this way:

P(-1.2

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

P(-1.2

3 0
4 years ago
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