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sveta [45]
3 years ago
14

A certain restaurant offers 8 different salads, 5 different main courses, 6 different desserts. If customers choose one salad, o

ne main course and two different desserts for their meal, how many different meals are possible?
A. 120
B. 240
C. 480
D. 600
E. 1200
Mathematics
1 answer:
Naddika [18.5K]3 years ago
8 0

Answer: E. 1200

Step-by-step explanation:

Given : A certain restaurant offers 8 different salads, 5 different main courses, 6 different desserts.

i.e. No. of choices for different salads= 8  -----(1)

No. of choices for different main courses = 5  ----(2)

No. of choices for different desserts = 6

When we choose two different desserts then we use permutation(repeatition not allowed) :

^6P_2=\dfrac{6!}{(6-2)!}=\dfrac{6\times5\times4!}{4!}=6\times5=30----(3)

[∵ No. of ways to choose r things out of n =^nP_r=\dfrac{n!}{(n-r)!} ]

If customers choose one salad, one main course and two different desserts for their meal , then By Fundamental principle of counting (Multiply (1) , (2) and (3)), the number of  different meals are possible :-

8\times5\times30\\\\=1200

Hence, the correct answer is E. 1200 .

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Answer:

a) P(X

And we can find this probability using the normal standard distribution or excel and we got:

P(z

b) P(X>40)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-43}{4.5})=P(z>3.78)

And we can find this probability using the complement rule and normal standard distribution or excel and we got:

P(z>3.78)=1-P(Z

c) z=0.674

And if we solve for a we got

a=43 +0.674*4.5=46.033

So the value of height that separates the bottom 75% of data from the top 25% is 46.033.  

Step-by-step explanation:

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".  

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

X \sim N(43,4.5)  

Where \mu=43 and \sigma=4.5

Part a

We are interested on this probability

P(X

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

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

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

P(X

And we can find this probability using the normal standard distribution or excel and we got:

P(z

Part b

P(X>40)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-43}{4.5})=P(z>3.78)

And we can find this probability using the complement rule and normal standard distribution or excel and we got:

P(z>3.78)=1-P(Z

Part c

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.25   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.674

And if we solve for a we got

a=43 +0.674*4.5=46.033

So the value of height that separates the bottom 75% of data from the top 25% is 46.033.  

8 0
3 years ago
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