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V125BC [204]
3 years ago
11

Suppose you can replace one number cube with a nonstandard number cube, where any of the numbers 1 through 6 can appear on multi

ple faces. How can you arrange the numbers on the nonstandard cube so that the mean of the rolls is the same as that of two standard number cubes, but the standard deviation is as large as possible? What is this value? Explain your thinking
Mathematics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

Replace 5 and 4 with 6s. Replace 2 and 3 with 1s. Then there will be 3 faces with 6 and 3 faces with 1.

Step-by-step explanation:

In order for the mean to remain unchanged, the sum of opposite faces must remain the same: 7. In order to have the standard deviation as large as possible, the largest and smallest possible numbers need to be used: 6 and 1.

Replacing 4 and 5 with 6s, and replacing 2 and 3 with 1s will accomplish your goal.

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Identify the type of sequence shown in the table below and select the appropriate response. n f(n) 1 540 2 500 3 460 4 420 5 380
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The answer is the first one; it is an arithmetic sequence with a common difference of -40.
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A dairy company gets milk from two dairies and then blends the milk to get the desired amount of butterfat. Milk from dairy I co
zubka84 [21]

Answer:

a) i The company should buy 40 gallons from dairy I and 60 gallons from dairy

ii) What is the maximum amount of​ butterfat? The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b.The excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

Step-by-step explanation:

a. How much milk from each supplier should the company buy to get at most 100 gallons of milk with the maximum amount of​ butterfat?

From the question, we are told that:

Milk from dairy I costs ​$2.40 per​ gallon, Milk from dairy II costs ​$0.80 per gallon.

Let's represent:

Number of gallons of Milk from dairy I = x

Number of gallons of Milk from dairy II = y

At most ​$144 is available for purchasing milk.

$2.40 × x + $0.80 × y = 144

2.40x + 0.80y = 144........ Equation 1

x + y = 100....... Equation 2

x = 100 - y

2.40(100 - y) + 0.80y = 144

240 - 2.4y + 0.80y = 144

-1.60y = 144 - 240

-1.6y = -96

y = -96/-1.6

y = 60

From Equation 2

x + y = 100....... Equation 2

x + 60 = 100

x = 100 - 60

x = 40

Therefore, since number of gallons of Milk from dairy I = x and number of gallons of Milk from dairy II = y

The company should buy 40 gallons from dairy I and 60 gallons from dairy

II. What is the maximum amount of​ butterfat?

From the question

Dairy I can supply at most 50 gallons averaging 3.9​% ​butterfat,

50 gallons = 3.9% butterfat

40 gallons =

Cross Multiply

= 40 × 3.9/50

= 3.12%

Dairy II can supply at most 90 gallons averaging 2.9​% butterfat.

90 gallons of milk = 2.9% butter fat

60 gallons of milk =

Cross Multiply

= 60 × 2.9%/90

=1.9333333333%

≈ 1.93%

The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b. The solution from part a leaves both dairy I and dairy II with excess capacity. Calculate the amount of additional milk each dairy could produce.

Excess capacity of Diary I =

50 gallons - 40 gallons = 10 gallons

Excess capacity of Diary II =

90 gallons - 60 gallons = 30 gallons

Therefore, the excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

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Answer:

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Answer:

g = - 27

Step-by-step explanation:

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