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Wittaler [7]
3 years ago
5

A seasonal index for a monthly series is about to be calculated on the basis of three years’ accumulation of data. The three pre

vious July values were 110, 150, and 130. The average over all months is 190. The approximate seasonal index for July is:
Mathematics
1 answer:
yarga [219]3 years ago
6 0

Answer:

0.684

Step-by-step explanation:

According to the scenario, computation of the given data are as follows

Seasonal index = Average value for July ÷ Average over all months

Where, Average value for July = ( 110 + 150 + 130 ) ÷ 3

= 390 ÷ 3 = 130

And, average over all months = 190

So by putting the value in the formula, we get

Seasonal index = 130 ÷ 190

= 0.684211 or 0.684

Hence, approximate seasonal index for July is 0.684.

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1.)Each sandwich costs $.02 per cubic inch to make. What is the cost of each sandwich?
kirza4 [7]
Part 1) <span>Each sandwich costs $.02 per cubic inch to make. What is the cost of each sandwich?

</span>volume for cube sandwich is 125 in³
volume for pyramid sandwich is 41.67 in³

cost of each cube sandwich=0.02*125-----> $2.5

cost of each  pyramid sandwich=0.02*41.67-----> $0.83

the answers Part 1) are
a) cost of each cube sandwich is $2.5
b) cost of each pyramid sandwich is $0.83

Part 2) <span>If each sandwich sells for $4.00 and Mr. Sammie sells 100 Cube sandwiches and 100 Pyramid Sandwiches, how much profit would Mr. Sammie make?
</span>
we know that
Profit=Sells-Costs

step 1
 find the sells
Sells=200*$4-----> $800

step 2
find the costs
Costs=100*$2.5+100*$0.83----> $333

step 3
find the profit
Profit=Sells-Costs
Profit=$800-$333-----> $467

the answer Part 2) is
the profit is $467
7 0
3 years ago
What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success​ p? C
charle [14.2K]

Answer:

(D)E[ X ] =np.

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success​ p,

f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq  x\leq n

E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}

Now,

x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)

Substituting,

E(X)=\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^x(1-p)^{n-x}

Factoring out the n and one p from the above expression:

E(X)=np\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^{x-1}(1-p)^{(n-1)-(x-1)}

Representing k=x-1 in the above gives us:

E(X)=np\sum_{k=0}^{n}n\left(\begin{array}{c}n-1\\k\end{array}\right)p^{k}(1-p)^{(n-1)-k}

This can then be written by the Binomial Formula as:

E[ X ] = (np) (p +(1 - p))^{n -1 }= np.

5 0
3 years ago
Match the pairs of equations that represent concentric circles. Tiles
Darina [25.2K]
I will explain you and pair two of the equations as an example to you. Then, you must pair the others.

1) Two circles are concentric if they have the same center and different radii.

2) The equation of a circle with center xc, yc, and radius r is:

(x - xc)^2 + (y - yc)^2 = r^2.

So, if you have that equation you can inmediately tell the coordinates of the center and the radius of the circle.

3) You can transform the equations given in your picture to the form (x -xc)^2 + (y -yc)^2 = r2 by completing squares.

Example:

Equation: 3x^2 + 3y^2 + 12x - 6y - 21 = 0

rearrange: 3x^2 + 12x + 3y^2 - 6y = 21

extract common factor 3: 3 (x^2 + 4x) + 3(y^2 -2y) = 3*7

=> (x^2 + 4x) + (y^2 - 2y) = 7

complete squares: (x + 2)^2 - 4 + (y - 1)^2 - 1 = 7

=> (x + 2)^2 + (y - 1)^2 = 12 => center = (-2,1), r = √12.

equation: 4x^2 + 4y^2 + 16x - 8y - 308 = 0

rearrange: 4x^2 + 16x + 4y^2 - 8y = 308

common factor 4: 4 (x^2 + 4x) + 4(y^2 -8y) = 4*77

=> (x^2 + 4x) + (y^2 - 2y) = 77

complete squares: (x + 2)^2  - 4 + (y - 1)^2 - 1 = 77

=> (x + 2)^2 + (y - 1)^2 = 82 => center = (-2,1), r = √82

Therefore, you conclude that these two circumferences have the same center and differet r, so they are concentric.
5 0
3 years ago
Is there anyone that can help me with this problem?
saul85 [17]

Answer:

your answer will be option 2

None of the choices are correct.

Step-by-step explanation:

have a nice day and good luck!!!

6 0
3 years ago
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ikadub [295]
The answer to this question is 70%
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3 years ago
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