Use the accompanying time series to complete parts a through c below. E Click the icon to view the time series data table. ..... a. Forecast the demand for each quarter in Year 4 using seasonal dummy variables and a best subsets regression. (Let the first dummy variable be equal to 1 for Quarter 2, and so on, following the order of the seasonal categories in the given table.) The demand forecast for the first quarter in Year 4 is 25.7 units. Time Series Data Table (Round to one decimal place as needed.) The demand forecast for the second quarter in Year 4 is units. (Round to one decimal place as needed.) Year 1 Demand 32 21 Year 3 Year 2 Demand Quarter Quarter Quarter Demand 31 17 27 15 26 24 22 4 13 4 10

Q: S ds J V3s? + 1

A: To evaluate the given integral

Q: 64y2+7=32

A: 64y2+7=32

Q: V-539 7. V11

A: To Determine:  Simplify

Q: Use the graph to solve the equation: x + 5 = 0.5(x + 3)² - 2 Answer

A:

Q: 2 +,

A:

Q: 2. K 5x+5 16 35

A:

Q: 4h-3=9

A: We have to find h

Q: 4w=25

A: topic - linear equation

Q: -12+(4• 3) (-) I 2-14 - 3)ENTER

A:

Q: Todt 5(-37+4)

A: Given,   ∫  -3t+410dt

Q: 2x=axz

A: Given equation is 2x=axz

Q: 2x Ax) = %3D 2 - 1

A:

Q: 3p - p

A: Here Given that, 3p-p

Q: -3w+5=23

A:

Q: 5×-6=37

A: Given,           5×-6=37

Q: -4d = -64

A: Given  -4d=-64

Q: -5 -2 1

A:

Q: S V25 – 4x²

A: For the solution follow the next step.

Q: -8 = 6+n

A:

Q: 2: Find: L-1[: 1 3 - 3]

A:

Q: 4т V17 3

A: Modulus function always gives positive quantity in result.

Q: 6+12=

A: Given,          6+12

Q: A+12=13

A: Given,   A+12=13     We will solve it for the value of A

Q: 4a5b• V18a3b6

A: We have to find

Q: 6+197

A: Given,          6+197

Q: 16+3•5

A: Given:  16+3*5

Q: S A+ AMQ for M.

A: NOTE: Refresh your page if you can't see any equations. .

Q: カメ Z

A: Find value of  2+2x4

Q: Find

A: We know that,                         ∫1xdx=lnx+ c We need to find  ∫4xdx

Q: 2+36=38

A: Given 2+ 36 = 38  Yes the given statement is true .

Q: dx 4е2я — 8ет + 5

A: This question can be solved using the concept of special integrals.

Q: 4h aー-5

A: We have to find out

Angles in Circles

Angles within a circle are feasible to create with the help of different properties of the circle such as radii, tangents, and chords. The radius is the distance from the center of the circle to the circumference of the circle. A tangent is a line made perpendicular to the radius through its endpoint placed on the circle as well as the line drawn at right angles to a tangent across the point of contact when the circle passes through the center of the circle. The chord is a line segment with its endpoints on the circle. A secant line or secant is the infinite extension of the chord.

Arcs in Circles

A circular arc is the arc of a circle formed by two distinct points. It is a section or segment of the circumference of a circle. A straight line passing through the center connecting the two distinct ends of the arc is termed a semi-circular arc.