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REY [17]
3 years ago
12

when planning for a party one caterer recommends the amount or meat be at least 2 pounds less than 1/3 the total number of guest

s. which graph represents this scenario?​

Mathematics
2 answers:
Darina [25.2K]3 years ago
6 0

Answer:

Its the 4th graph.

Step-by-step explanation:

IgorLugansk [536]3 years ago
3 0

Answer:

Let x be the number of guest and y be the quantity of meat,

According to the question,

y\geq \frac{x}{3}-2

Since, the related equation of the above inequality,

y=\frac{x}{3}-2

Having x-intercept = (6,0),

y-intercept = (0,-2)

Also,'≥' shows the solid line,

Now, 0 ≥ 0/3 - 2  ( true )

Hence, the shaded region of above inequality will contain the origin,

Therefore, by the above information we can plot the graph of the inequality ( shown below ).

You might be interested in
Irene made a mistake when she solved this inequality:
just olya [345]

Answer:

she messed up on step two because she has to subtract 10 from both sides

Step-by-step explanation:

step 1: -6(x+3)+10<-2

step2:-6(x+3)+10-10<-2-10

step3: -6(x+3)<-12

step4: (-6)(x+3)(-1)≥(-12)(-1)

step5:6(x+3)>12

step6:divide both sides by 6

step7:simplify and subtract 3 from both sides and then simplify again

3 0
3 years ago
A deck of 52 cards contains 12 picture cards. If the 52 cards are distributed in a random manner among four players in such a wa
Mkey [24]

Answer:

The probability that each player will receive three picture cards = 0.0324

Step-by-step explanation:

As given,

A deck of 52 cards contains 12 picture cards

Remaining card = 52 - 12 = 40

So,

Total number of ways in which 12 picture card is distributed = \frac{12!}{3! 3! 3! 3!}

Now,

The Total number of ways in which Remaining cards are distributed = \frac{40!}{10! 10! 10! 10!}

So,

Total number of ways of getting 3 picture card and remaining card = \frac{12!}{3! 3! 3! 3!}× \frac{40!}{10! 10! 10! 10!}

= \frac{12! 40!}{(3!)^{4} (10!)^{4}  }

Now,

Total number of ways to distribute 52 cards so that each people get 13 card = \frac{52!}{13! 13! 13! 13!} = \frac{52!}{ (13!)^{4} }

∴ The probability = \frac{\frac{12! 40!}{(3!)^{4} (10!)^{4}  }}{\frac{52!}{(13!)^{4} }}

                            = \frac{12! 40!}{(3!)^{4} (10!)^{4}  }×\frac{(13!)^{4} }{ 52! }

                           = \frac{12! 40!}{(3!)^{4} (10!)^{4}  }×\frac{(13.12.11.10!)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41.40! }

                           = \frac{12!}{(3!)^{4}   }×\frac{(13.12.11)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}

                           = \frac{479,001,600}{(6)^{4}   }×\frac{(1716)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}

                           = 0.0324

∴ we get

The probability that each player will receive three picture cards = 0.0324

6 0
3 years ago
A store sells both cold and hot beverages. Cold beverages, c, cost $1.50, while hot beverages, h, cost $2.00. On Saturday, drink
Marina86 [1]
2h + 4h  (1.5c) = 360 so option b is correct.
3 0
3 years ago
Read 2 more answers
Suppose that your boss must choose three employees in your office to attend a conference in Jamaica. Because all 17 of you want
Pani-rosa [81]

Answer:

P(You\ n\ Anna\ n\ Kevin) = \frac{1}{4913}

Step-by-step explanation:

Given

Total\ Employees = 17

To\ Select = 3

Required

Determine the probability of selecting You, Anna and Kevin

Assuming the selection process is fair;

Each employee has a probability of;

Probability = \frac{1}{17}

So:

P(You\ n\ Anna\ n\ Kevin) = P(You) * P(Anna) * P(Kevin)

P(You\ n\ Anna\ n\ Kevin) = \frac{1}{17} * \frac{1}{17} * \frac{1}{17}

P(You\ n\ Anna\ n\ Kevin) = \frac{1}{4913}

5 0
3 years ago
5. A square box has a volume of 64
oksano4ka [1.4K]

Answer:

a.

Step-by-step explanation:

Volume of square = length³

64 = length ³

Taking cube root in both sides

We'll get

Length = 4

Now perimeter of square = 4×L

= 4×4

= 16

5 0
3 years ago
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