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ss7ja [257]
3 years ago
5

The owner of a new restaurant is ordering tables and chairs. He wants to have only tables for 2 and tables for 4. The total numb

er of people that can be seated in the restaurant is 120.
a. Find some possible combinations of 2-seat tables and 4-seat tables that will seat 120 customers.
Mathematics
1 answer:
gayaneshka [121]3 years ago
4 0

Answer:

5 points for what

Step-by-step explanation:

HeY

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What is logs(4.7)+logs 2 written as a single log?
Ipatiy [6.2K]

ANSWER

log_{s}(56)

EXPLANATION

The given logarithmic expression is:

log_{s}(4 \times 7)  +  log_{s}(2)

Recall and use the product rule of logarithm

log_{a}(b)  +  log_{a}(c)  =  log_{s}(bc)

We apply this rule to obtain,

log_{s}(4 \times 7)  +  log_{s}(2)  = log_{s}(4 \times 7 \times 2)

We multiply out the argument to get;

log_{s}(4 \times 7)  +  log_{s}(2)  = log_{s}(56)

The correct answer is

log_{s}(56)

7 0
4 years ago
Find area of circle ​
Gre4nikov [31]

Answer and Step-by-step explanation:

Plug in 4 for r, since 4 is half of 8, and 8 is the diameter, and the radius is half of the diameter.

We get the answer to be 16\pi, or 50.2655.

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3 years ago
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How do you simplify -50 1/2+ 12.3
marishachu [46]

Answer:

-38.2

Step-by-step explanation:

6 0
3 years ago
What is 56/30 in its simplest form?
lbvjy [14]

Answer:

1\frac{13}{15}

Step-by-step explanation:

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2 years ago
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A box contains four red balls and eight black balls. Two balls are randomly chosen from the box, and are not
castortr0y [4]

P(B) = 8/12

P(R | B) = 4/11

P(B ∩ R) = 8/33

The probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

<em><u>Solution:</u></em>

<em><u>The probability is given as:</u></em>

Probability = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}

Given that,

A box contains four red balls and eight black balls

Red = 4

Black = 8

Total number of possible outcomes = 12

Let event B be choosing a black ball first and event R be choosing a red ball second.

<h3><u>Find P(B)</u></h3>

P(B) = \frac{8}{12}

<h3><u>Find P(B n R)</u></h3>

P(B n R) = P(B) \times P(R)\\\\P(B n R) = \frac{8}{12} \times \frac{4}{11}\\\\P(B n R) = \frac{8}{33}

<h3><u>Find </u><u> P(R | B)</u></h3><h3>P(R | B) = \frac{P(R n B)}{P(B)}\\\\P(R | B) = \frac{\frac{8}{33}}{\frac{8}{12}}\\\\P(R | B) = \frac{8}{33} \times \frac{12}{8}\\\\P(R | B) = \frac{4}{11}</h3>

<em><u>The probability that the first ball chosen is black and the second ball chosen is red is about percent</u></em>

\frac{8}{33} \times 100 = 0.24 \times 100 = 24 \%

Thus the probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

4 0
4 years ago
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