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Dmitrij [34]
3 years ago
11

Which of the following is an example of a causal regression analysis forecast? To forecast how many hamburgers to sell tomorrow

based on tomorrow's forecasted weather To forecast how many hamburgers to sell tomorrow based on one's gut feeling To forecast how many hamburgers to sell tomorrow based on next year's budget limit To forecast how many hamburgers to sell tomorrow based on past period sales figures
Mathematics
2 answers:
Ira Lisetskai [31]3 years ago
7 0

Answer: To forecast how many hamburgers to sell tomorrow based on past period sales figures.

Step-by-step explanation:Regression analysis forecast is a term used to describe an analysis which focuses on forecasting and predicting based on  the available data.

Forecasting is a term used in economics to predict the possible outcomes of a particular situation based on previous knowledge or experiences.

An example of a causal regression analysis forecast is making a forecast of how many products to sell tomorrow using already available past sales details.

labwork [276]3 years ago
3 0

Answer:

To forecast how many hamburgers to sell tomorrow based on past period sales figures.

Step-by-step explanation:

Regression analysis explains predicting sales based on finding a relationship between past sales and one or more independent variables, such as population or income.

You might be interested in
-5x - 35= -15<br> -5x = 25<br> X =
Zigmanuir [339]
X= -5 because you divide -5x by -5 to get x alone so you have to 25/-5 which is -5
3 0
3 years ago
Consider the equation and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”. For example, “(0, 3) R (0, 2)”
Leviafan [203]

Answer:

The equation determine a relation between x and y

x = ± \sqrt{1-(y-2)^{2}}

y = ± \sqrt{1-x^{2}}+2

The domain is 1 ≤ y ≤ 3

The domain is -1 ≤ x ≤ 1

The graphs of these two function are half circle with center (0 , 2)

All of the points on the circle that have distance 1 from point (0 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

 R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

 of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* <em>Lets solve the problem</em>

- The equation of a circle of center (h , k) and radius r is

  (x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒  (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

∴ The equation determine a relation between x and y

* Lets prove that x=g(y)

- To do that find x in terms of y by separate x in side and all other

  in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

∴ x² = 1 - (y - 2)²

- Take square root for both sides

∴ x = ± \sqrt{1-(y-2)^{2}}

∴ x = g(y)

* Lets prove that y=h(x)

- To do that find y in terms of x by separate y in side and all other

  in the other side

∵ x² + (y - 2)² = 1

- Subtract x² from both sides

∴ (y - 2)² = 1 - x²

- Take square root for both sides

∴ y - 2 = ± \sqrt{1-x^{2}}

- Add 2 for both sides

∴ y = ± \sqrt{1-x^{2}}+2

∴ y = h(x)

- In the function x = ± \sqrt{1-(y-2)^{2}}

∵ \sqrt{1-(y-2)^{2}} ≥ 0

∴ 1 - (y - 2)² ≥ 0

- Add (y - 2)² to both sides

∴ 1 ≥ (y - 2)²

- Take √ for both sides

∴ 1 ≥ y - 2 ≥ -1

- Add 2 for both sides

∴ 3 ≥ y ≥ 1

∴ The domain is 1 ≤ y ≤ 3

- In the function y = ± \sqrt{1-x^{2}}+2

∵ \sqrt{1-x^{2}} ≥ 0

∴ 1 - x² ≥ 0

- Add x² for both sides

∴ 1 ≥ x²

- Take √ for both sides

∴ 1 ≥ x ≥ -1

∴ The domain is -1 ≤ x ≤ 1

* The graphs of these two function are half circle with center (0 , 2)

* All of the points on the circle that have distance 1 from point (0 , 2)

8 0
3 years ago
Type the correct answer in each box
Luba_88 [7]

Answer:

We know that the equation of the circle in standard form is equal to <em>(x-h)² + (y-k)² = r²</em> where (h,k) is the center of the circle and r is the radius of the circle.

We have x² + y² + 8x + 22y + 37 = 0, let's get to the standard form :

1 - We first group terms with the same variable :

(x²+8x) + (y²+22y) + 37 = 0

2 - We then move the constant to the opposite side of the equation (don't forget to change the sign !)

(x²+8x) + (y²+22y) = - 37

3 - Do you recall the quadratic identities ? (a+b)² = a² + 2ab + b². Now that's what we are trying to find. We call this process <u><em>"completing the square"</em></u>.

x²+8x = (x²+8x + 4²) - 4² = (x+4)² - 4²

y²+22y = (y²+22y+11²)-11² = (y+11)²-11²

4 - We plug the new values inside our equation :

(x+4)² - 4² + (y+22)² - 11² = -37

(x+4)² + (y+22)² = -37+4²+11²

(x+4)²+(y+22)² = 100

5 - We re-write in standard form :

(x-(-4)²)² + (y - (-22))² = 10²

And now it is easy to identify h and k, h = -4 and k = - 22 and the radius r equal 10. You can now complete the sentence :)

5 0
3 years ago
Suppose that X is a subset of Y. Let p be the proposition ‘x is an element ofX’ and let q be the proposition ‘x is an element of
Genrish500 [490]

Answer:

See answer below

Step-by-step explanation:

The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.

i) x∈AnB  if and only (if and only if means that both implications hold) x∈A and x∈B if and only if x∈A and x∉B^c (because B^c is the set of all elements that do not belong to X) if and only if x∈A\B^c. Then, if x∈AnB  then x∈A\B^c, and if x∈A\B^c then x∈AnB. Thus both sets are equal.

ii) (I will abbreviate "if and only if" as "iff")

x∈A∪(B\A) iff x∈A or x∈B\A iff x∈A or x∈B and x∉A iff x∈A or x∈B (this is because if x∈B and x∈A then x∈A, so no elements are lost when we forget about the condition x∉A) iff x∈A∪B.

iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).

iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).

8 0
3 years ago
Solve for k.<br> 6k - 8 = 4K + 15
Brut [27]

Hi,

Answer:

k = \frac{23}{2}

Step-by-step explanation:

Subtract 4k from both sides

6k - 4k = 2k

2k - 8 = 15

Add 8 on both sides (you want to get rid of the 8 in order to leave the k alone)

2k = 23

k = 23/2

Have a good day!

4 0
3 years ago
Read 2 more answers
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