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Usimov [2.4K]
3 years ago
12

You are selecting meals for a school event and you are planning on at least 240 students at this event. The chicken meal will co

st $5 and the beef meal will cost $8 per serving. Your total budget is $1200.00. Select a system of linear inequalities that shows the various numbers of Chicken and Beef meals you can prepare. Let x = number of chicken meals and let y = number of beef meals.
Mathematics
1 answer:
den301095 [7]3 years ago
3 0

Answer:

x + y ≥ 240

5x + 8y ≤ 1200

Step-by-step explanation:

Assume;

Number of chicken meals = x

Number of Beef meals = y

Total budget = $1200

Find:

Inequality

Computation:

Number of chicken meals + Number of Beef meals  ≥ Total students

x + y ≥ 240

Total cost = (Number of chicken meals)(Cost of chicken meal) + (Number of Beef meals)(Cost of beaf meal)

5x + 8y ≤ 1200

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Answer:

See explanation

Step-by-step explanation:

Given

See attachment for proper presentation of question

Required

Mean and Range

To do this, we simply calculate the mean and the range of each row.

\bar x = \frac{\sum x}{n} ---- mean

Where:

n = 4 ---- number of rows

R = Highest - Lowest --- range

So, we have:

Sample 1

\bar x_1 = \frac{1027+ 994 +977 +994 }{4}

\bar x_1 = 998

R_1 = 1027- 994

R_1 = 33

Sample 2

\bar x_2 = \frac{975 +1013 +999 +1017}{4}

\bar x_2 = 1001

R_2 =  1017 - 975

R_2 = 42

Sample 3

\bar x_3 = \frac{988 +1016 +974 +997}{4}

\bar x_3 = 993.75

R_3 = 1016-974

R_3 = 42

Sample 4

\bar x_4 = \frac{998 +1024 +1006 +1010}{4}

\bar x_4 = 1009.5

R_4 = 1024 -998

R_4 = 26

Sample 5

\bar x_5 = \frac{990 +1012 +990 +1000}{4}

\bar x_5 = 998

R_5 = 1012 -990

R_5 = 22

Sample 6

\bar x_6= \frac{1016 + 998 +1001 +1030}{4}

\bar x_6= 1011.25

R_6= 1030-998

R_6= 32

Sample 7

\bar x_7 = \frac{1000 +983 +979 +971}{4}

\bar x_7 = 983.25

R_7 = 1000-971

R_7 = 29

Sample 8

\bar x_8 = \frac{973 +982 +975 +1030}{4}

\bar x_8 = 990

R_8 = 1030-973

R_8 = 57

Sample 9

\bar x_9 = \frac{992 +1028 +991 +998}{4}

\bar x_9 = 1002.25

R_9 = 1028 -991

R_9 = 37

Sample 10

\bar x_{10} = \frac{997 +1026 +972 +1021}{4}

\bar x_{10} = 1004

R_{10} = 1026 -972

R_{10} = 54

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\bar x_{11} = 1007.75

R_{11} = 1028 -990

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\bar x_{12} = 996.25

R_{12} = 1021 -970

R_{12} = 51

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R_{13} =1027 -993

R_{13} =34

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\bar x_{14} = \frac{1022 +981 +1014 +983}{4}

\bar x_{14} = 1000

R_{14} = 1022 -981

R_{14} = 41

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\bar x_{15} = \frac{977 +993 +986 +983}{4}

\bar x_{15} = 984.75

R_{15} = 993-977

R_{15} = 16

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Answer:

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Step-by-step explanation:

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2.75 x 3 = 8.25

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Answer:

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