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Ilia_Sergeevich [38]
3 years ago
7

Trey's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Trey $4.80 per pound, and type B

coffee costs $5.85 per pound. This month's blend used three times as many pounds of type B coffee as type A, for a total cost of $603.45 . How many pounds of type A coffee were used?
Mathematics
1 answer:
boyakko [2]3 years ago
4 0

Let x = the number of pounds of Type A coffee.

 

We know that this month's blend used 3 times as many pounds of type B coffee as type A coffee.

Total pounds of coffee =  x + 3x

 

The total cost is:

(cost for Type A)(x) + (cost for Type B)(3x) = $603.45

$4.80x + $5.85(3x) = $603.45

$4.80x + $17.55x = $603.45

$22.35x = $603.45

x = 603.45/22.35 = 27 pounds


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Answer:

Option B

Step-by-step explanation:

Here we have to apply " combination and permutation. " It is given that the drama club had to choose three booths from a selection of 9, considering the possible ways to choose so. This is a perfect example of combination. In nCr, n corresponds to 9, respectively r corresponds to 3.

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What is an example of when you would want consistent data and, therefore, a small standard deviation?
steposvetlana [31]

Answer:

12.1, 12.3,12.4,12.5,12.3,12.1,12.2

\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271

And for the standard deviation we can use the following formula:

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And after replace we got:

s = 0.1496

And as we can ee we got a small value for the deviation <1 on this case.

Step-by-step explanation:

For example if we have the following data:

12.1, 12.3,12.4,12.5,12.3,12.1,12.2

We see that the data are similar for all the observations so we would expect a small standard deviation

If we calculate the sample mean we can use the following formula:

\bar X=\frac{\sum_{i=1}^n X_i}{n}

And replacing we got:

\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271

And for the standard deviation we can use the following formula:

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And after replace we got:

s = 0.1496

And as we can ee we got a small value for the deviation <1 on this case.

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