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jek_recluse [69]
3 years ago
11

A restaurant menu has five kinds of soups, seven kinds of main courses, six kinds of desserts, and five kinds of drinks. If a cu

stomer randomly selects one item from each of these four categories, how many different outcomes are possible
Mathematics
2 answers:
xxMikexx [17]3 years ago
5 0

Answer:

1050

Explanation:

Multiply each of the choices together.

5 x 7 x 6 x 5 = 1050

ryzh [129]3 years ago
3 0

Answer: 1050

Step-by-step explanation:

Number of combinations of selecting r things out of n = ^nC_r=\dfrac{n!}{r!(n-r)!}

such that ^nC_1=n

Given: A restaurant menu has 5 kinds of soups, 7kinds of main courses, 6 kinds of desserts, and 5 kinds of drinks.

If a customer randomly selects one item from each of these four categories, then by fundamental counting principle , the number of different outcomes are possible = ^5C_1\times \ ^7C_1\times\ ^6C_1\times\ ^5C_1 =5\times7\times6\times5=1050

hence, total number of outcomes = 1050

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Answer:

a) a_n=3\,n-11

b) a_{20}=49

c) term number 17 is the one that gives a value of 40

Step-by-step explanation:

a)

The sequence seems to be arithmetic, and with common difference d = 3.

Notice that when you add 3 units to the first term (-80, you get :

-8 + 3 = -5

and then -5 + 3 = -2 which is the third term.

Then, we can use the general form for the nth term of an arithmetic sequence to find its simplified form:

a_n=a_1+(n-1)\,d

That in our case would give:

a_n=-8+(n-1)\,(3)\\a_n=-8+3\,n-3\\a_n=3n-11

b)

Therefore, the term number 20 can be calculated from it:

a_{20}=3\,(20)-11=60-11=49

c) in order to find which term renders 20, we use the general form we found in step a):

a_n=3\,n-11\\40=3\,n-11\\40+11=3\,n\\51=3\,n\\n=\frac{51}{3} =17

so term number 17 is the one that renders a value of 40

5 0
3 years ago
Can anyone help me on my HW i can't seem to find the answer to it and when i do it doesn't make sence
Nadusha1986 [10]
0.75x-18.5=0.65x
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To checkt:
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