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Ghella [55]
2 years ago
8

Alejandra's Tapas Bar offers a menu consisting of $9$ savory and $5$ sweet dishes. You can also get a mix-and-match plate consis

ting of two different dishes on the menu. How many different mix-and-match plates can you get consisting of two savory dishes?
Mathematics
1 answer:
Mariana [72]2 years ago
7 0

Answer:

45

Step-by-step explanation:

Given that the number of savory dishes is 9 and the number of sweet dished is 5.

Denoting all the 9 savory dishes by p_1, p_2,...,p_9, and all the sweet dishes by q_1,q_2,...,q_5.

The possible different mix-and-match plates consisting of two savory dishes are as follows:

There are 9 plates with q_1 from sweet plates which are (q_1, p_1), (q_1, p_2), ..., (q_1,p_9).

There are 9 plates with q_2 from sweet plates which are  (q_2, p_1), (q_2, p_2), ..., (q_2,p_9).

Similarly, there are 9 plated for each q_3, q_4 and q_5.

Hence, the total number of the different mix-and-match plates consisting of two savory dishes

= 9+9+9+9+9= 9\times5=45

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Answer:

26/75

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4 / 15 / 10 / 13

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= 26/75  answer

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2 years ago
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abruzzese [7]
Given data :
a₃ = 9/16
aₓ = -3/4 · aₓ₋₁
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We know that aₓ = -3/4 · aₓ₋₁
So,
a₃ = -3/4 · a₃₋₁
a₃ = -3/4 · a₂
9/16 = -3/4 · a₂
a₂ = 9/16 × -4/3
a₂ = -36/48
a₂ = -3/4

Again,
aₓ = -3/4 · aₓ₋₁
a₄ = -3/4 · a₄₋₁
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a₄ = -3/4 · 9/16
a₄ = -27/64
a₄ = -27/64

For a₅,
aₓ = -3/4 · aₓ₋₁
a₅ = -3/4 · a₅₋₁
a₅ = -3/4 · a₄
a₅ = -3/4 × -27/64
a₅ = 81/256

For a₆,
aₓ = -3/4 · aₓ₋₁
a₆ = -3/4 · a₆₋₁
a₆ = -3/4 · a₅
a₆ = -3/4 × 81/256
a₆ = -243/1024

For a₇,
aₓ = -3/4 · aₓ₋₁
a₇ = -3/4 · a₇₋₁
a₇ = -3/4 · a₆
a₇ = -3/4 × -243/1024
a₇ = 729/4096
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2 years ago
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Step-by-step explanation:

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Answer:

<u>option 1</u>

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