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Ad libitum [116K]
3 years ago
5

Help plsss!!!

Mathematics
1 answer:
Kisachek [45]3 years ago
4 0

Answer:

A and C

Step-by-step explanation:

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What is the simple interest on a loan of $16 500 for 2 years at 8 per year ?
Helen [10]

Answer:

$2 640

Step-by-step explanation:

Principal = $16 500; Time = 2 years Rate = 8%

Simple Interest = P × R × T/100

= $16 500 × 8 × 2/100

=$264 000/100

Simple Interest = $2 640

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Jose wants to find the perimeter of triangle ABC. He uses the distance formula to determine the length of AC. Finish Jose’s calc
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What's the probability you roll the same number twice on a regular dice?
jekas [21]
1/6 because the first number that you roll doesn't matter so it depends on the 2nd roll, in which the probability of rolling a specific number is 1/6
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4 years ago
The sum of the 3rd and 7th terms of an A.P. is 38, and the 9th term is 37. Find the A.P.
ElenaW [278]

Answer:

The AP is 1, 11/2, 10, 29/2, 19, ....

Step-by-step explanation:

Let the first term be a and d be the common difference of the arithmetic progression.

ATQ, a+2d+a+6d=38, 2a+8d=38 and a+8d=37. Solving this, we will get a=1 and d=9/2. The AP is 1, 11/2, 10, 29/2, 19, ....

4 0
3 years ago
URGENT please help!!! I need a explanation please!! You own a restaurant and can reopen as long as people seated at a table are
hram777 [196]

Answer:

1) Maximum number of tables that can be kept in the dinning room is 10 tables

2)The maximum number of customers, given 6 people per table is 60 customers

Step-by-step explanation:

The parameters given are;

Distance between people seated at different tables = 6 ft

Dimension of table = 2 m by 1.5 m

Number of people at a table = 6 people

Dimension of dinning;

Width = 12 m

Length = 20 m

Dimension of entrance;

Width = 20/21*30  m

Length = 10 m

With 6 meters between tables and a table with of 1.5, we have for the arrangement in the question;

3 tables = 4.5 m

Distance between = 2 × 6 = 12

4.5 + 12 = 16 m (More room required)

With two tables, we have;

Width of tables = 2×1.5 = 3 m

Distance between = 6 m

Dining room width required = 3 + 6 = 9 m

Therefore, the maximum tables in each row = 2

Given that the dining room area extends to the entrance area, total length = 30 m.

With 4 tables we have;

Length = 4 × 2 + 6 × 3 = 26

While on the entrance side of the dinning room area which is 20 m, we have 3 tables;

Length = 3 × 2 + 6 × 2 = 18

Therefore, both arrangements are acceptable;

Just on entering by the right the length = 20/21×30 m

Therefore, with 3 tables will required number due to proximity wit the door and tables arranged along the right wall of the dinning

1) Maximum number of tables that can be kept in the dinning room = 4 + 3 + 3 = 10 tables

2)The maximum number of customers, given 6 people per table = 6×10 = 60 customers.

4 0
4 years ago
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