Select the correct answer from each drop-down menu.

Igoryamba

Any thing equal to or greater than two thirds (2/3)

7 0

3 years ago

a catering service offers 6 appetizers, 10 main courses, and 8 desserts. A banquet committee is to select 2 appetizers, 3 main c

S_A_V [24]

There are two different answers that you could be looking for.
You might be asking how many different meals can be served at the banquet,
or you might be asking literally how many 'ways' there are to put meals together.

I'm going to answer both questions. Here's how to understand the difference:

Say you have ten stones, and you tell me "I'll let you pick out two stones
and take them home. How many ways can this be done ?"

For my first choice, I can pick any one of 10 stones. For each of those . . .
I can pick any one of the 9 remaining stones for my second choice.
So the total number of 'ways' to pick out two stones is (10 x 9) = 90 ways.

But let's look at 2 of those ways:
 -- If I pick stone-A first and then pick stone-G, I go home with 'A' and 'G'.
 -- If I pick stone-G first and then pick stone-A, I still go home with 'A' and 'G'.
There are two possible ways to pick the same pair.
In fact, there are two possible ways to pick every pair.
So there are 90 ways to pick a pair, but only 45 different pairs.

That's the reason for the difference between the number of ways the
committee can make their selections, and the number of different meals
they can put together for the banquet.

So now here's the answer to the question:

-- Two appetizers can be selected in (6 x 5) = 30 ways.
(But each pair can be selected in 2 of those ways,
so there are only 15 possible different pairs.)

-- Three main courses can be selected in (10 x 9 x 8) = 720 ways.
(But each trio can be selected in 3*2=6 of those ways,
so there are only 120 possible different trios.)

-- Two desserts can be selected in (8 x 7) = 56 ways.
(But each pair of them can be selected in 2 of those ways,
so there are only 28 possible different pairs.)

-- The whole line-up can be selected in (30 x 720 x 56) = 1,209,600 ways.

But the number of different meals will be (30 x 720 x 56) / (2 x 6 x 2) =

 (15 x 120 x 28) = 50,400 meals.

5 0

3 years ago

K 271° (3х – 4), (x + 60° M N L

Ainat [17]

Have a picture of the problem?

6 0

3 years ago

Read 2 more answers

Help me plez........

Vika [28.1K]

Answer:

=12