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Afina-wow [57]
3 years ago
8

You’ve been asked to order cake for your school dance. Here’s a formula for the scenario: z=xa/y

Mathematics
1 answer:
olga nikolaevna [1]3 years ago
7 0

Answer:

The number of cake for order is 15 cakes .

Step-by-step explanation:

Given as :

Total number of people to attend the dance = x = 60

The number of slices each person will eat = A = 2 slices

The number of slices per cake = y = 8 slices

Let The number of cake for order = z

<u>Now, According to question</u>

Total slices of cake required for 60 people when each people eat 2 slices = total number of people × number of slices each person will eat

i.e Total slices of cake required for 60 people when each people eat 2 slices = x × A

Or, Total slices of cake required for 60 people when each people eat 2 slices = 60 × 2 slices

Or,Total slices of cake required for 60 people when each people eat 2 slices = 120 slices

<u>Again</u>

∵ number of slices per cake = 8

∴ The number of cake for order = \dfrac{\textrm Total slices of cake required for 60 people when each people eat 2 slices}{\textrm number of slices per cake }

Or, z = \frac{x\times A}{y}

i.e z = \frac{120}{8}

∴   z = 15 cakes

So,The number of cake for order = z = 15 cakes

Hence,The number of cake for order is 15 cakes . Answer

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ZanzabumX [31]

Answer:

P(0.7

And we can find this probability with the following difference:

P(0.7

We can use the following commands on the ti 84

2nd>VARS>DISTR

And then we look for normalcdf and we input this:

normalcdf(0.7,1.4,0,1)

The other possible code would be:

normalcdf(-1000,1,4,0,1)-normalcdf(-1000,0.7,0,1)

And we got:

P(0.7

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

For this case we want to find this probability:

P(0.7

And we can find this probability with the following difference:

P(0.7

We can use the following commands on the ti 84

2nd>VARS>DISTR

And then we look for normalcdf and we input this:

normalcdf(0.7,1.4,0,1)

The other possible code would be:

normalcdf(-1000,1,4,0,1)-normalcdf(-1000,0.7,0,1)

And we got:

P(0.7

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Step-by-step explanation:

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