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Gelneren [198K]
4 years ago
6

Is for over two and 20/6 proportions

Mathematics
2 answers:
DedPeter [7]4 years ago
7 0
To find out if fractions are proportional you cross multiply.

4       20
--  x   ---
2        6

4 x 6 = 24

2 x 20 = 40

If they are alike then they are proportional, if they are not then they are NOT proportional.
ad-work [718]4 years ago
4 0
No they're not proportional, and it's four*.
4/2 = 2
20/6 = 3.333333333334 or 3 1/3
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Y_Kistochka [10]

Answer: 153

Step-by-step explanation:

When we have a group of N elements, the total number of combinations of K elements ( K ≤ N) is:

C (N, K) = \frac{N!}{(N-k)!*K!}

Where N! = N*(N - 1)*(N - 2)*...*2*1

In this case, we have a group of 18 people (then N = 18) and we want to see how many different combinations of 2 we can make (K = 2).

Using the above equation we get:

C(18, 2) = \frac{18!}{(18 - 2)!*2!}  = \frac{18!}{16!*2!} = \frac{18*17}{2}  = 9*17 = 153

There are 153 different ways in which the manager can do this.

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3 years ago
What is the mean of the normal distribution shown below? -1 0 1 2
Alisiya [41]

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Should be 1/2, or 0.5

Step-by-step explanation:

The mean is the average, so we divide the sum of the numbers(-1+0+1+2=2) by the quantity(4).

6 0
3 years ago
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alukav5142 [94]

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5 0
3 years ago
Suppose you are starting your own company selling chocolate covered strawberries. You decide to sell the milk chocolate covered
lidiya [134]

Answer:

Step-by-step explanation:

Let x_m and x_w be the production level of milk and white chocolate-covered strawberries respectively. According to the given data, we know the total profit will be

P(x_m,x_w)=$2.25x_m+$2.50x_w

The restrictions can be written as

x_m+x_w \leq  800

x_w\leq x_m/2

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All the restrictions can be plotted in the same graph to find the feasible region where all of them are met. The graph is shown in the image below

The optimal solution will be the level of production such that

* All restrictions are met

* The total profit is maximum

The optimal level of production can be found in (at least) one of the vertices of the feasible region. We'll try each one as follows

P(0,0)=0

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P(600,200)=$2.25 (600)+$2.50 (200) = $1850

P(800,0)=$2.25 (800)+$2.50 (0) = $1800

We must produce 600 milk chocolate-covered strawberries and 200 white chocolate-covered strawberries to have a maximum profit of $1850/month

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RUDIKE [14]
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