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Ad libitum [116K]
3 years ago
12

An old-fashioned Chinese restaurant offers a family dinner where you get to choose one dish from “column A” (which has 8 dishes)

, one dish from “column B” (which has 10 dishes) and one dish from “column C” (which has 5 dishes). How many different family dinners can be chosen?
Mathematics
2 answers:
erastovalidia [21]3 years ago
6 0
23 differrent dinners

adoni [48]3 years ago
3 0

Answer:

There are 400 ways to choose the dinner.

Step-by-step explanation:

Since we have to choose one dish from each column and we have three columns A ,B and C .

Number of choices for Column A = 8

Number of choices for column B = 10

Number of choices for column C = 5 dishes

Number of different family dinners can be chosen = 8x10 x5

                                                                                     = 400 ways

                                                [Using fundamental principle of Multiplication]

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Which of the following has a value between 10/3 and 11 / 3
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3 years ago
Maria and Katy have string Maria's is 6 cm longer that Katy's. When they put both of the strings together the total is 84 How lo
butalik [34]

Answer:

Maria's string is 45 cm long and Katy's string is 39 cm long

Step-by-step explanation:

Let's call the size of Maria's string X, and the size of Katy's string Y.

So, using the information from the problem, we can formulate a system of equations:

If Maria's string is 6 cm longer than Katy's, so:

X = Y + 6

If the strings together have 84 cm, so:

X + Y = 84

The first equation can be rewritten as X - Y = 6

Summing up both equation, we have that:

2X = 90

X = 45

Using X value in the first equation, we have that:

45 = Y + 6

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7 0
3 years ago
0.39<br> which is best buy<br> 0.41<br> 0.4.
Nana76 [90]

Answer:

0.39 because it's the cheapest / least

Step-by-step explanation:

Ok so we are trying to find the least. So first line up your decimals as shown below

<u>Tens</u>          <u>Tenths</u>      <u>Hundredths</u>

<h3>   0      .        3              9</h3><h3>---------------------------------------</h3><h3>   0      .        4              1</h3><h3>---------------------------------------</h3><h3>   0      .        4              0 ---> Imaginary 0</h3><h3 />

Now fill in any empty spaces with a 0 as shown in the decimal 0.4, your decimal will still have the same amount because 0.4 = 0.40 = 0.400 no matter how many 0's you put in the decimal.

When your done with placing 0's, look at the numbers in the first column, the Tens column. As you can see we have the numbers 0, 0, 0. They're all the same so we don't have to worry about that. Now lets go to the next column, the Tenths column. We have the numbers 3, 4, 4. Remember, we are finding the least amount. we know 3 is less than 4 so the decimal with the 3 in the Tenths place is the best buy. If these numbers were the same we would go on to the next place, the Hundredths place. If 2 of the decimals had a 3 in the Tenths place, you would eliminate the decimal with no 3 in the Tenths place and go on to the next place using only the decimals you have left.

So, <em>0.39</em> is the best buy.

Hope this helps you and makes Sense! :)

This is NOT copy pasted, I actually wrote all of this so that the person reading this answer can understand what I'm trying to explain.

Thank you, and have a Great Day! ;)

5 0
3 years ago
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