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solniwko [45]
3 years ago
15

After a new product is launched the cumulative sales S(t) (in $1000) t weeks after launch is given by:

Mathematics
1 answer:
LenaWriter [7]3 years ago
8 0

Answer:

$17.750 ; 15.979 ; 72

Step-by-step explanation:

Given that :

Cummulative sales, S(t) is represented by the equation :

S(t) = 72/(1 + 9e^-0.36t)

Cummulative sales after 3 weeks :

Put t = 3 in the equation, as t = time after launch

S(3) = 72/(1 + 9e^-0.36(3))

S(3) = 72 / (1 + 9e^-1.08)

S(3) = 72 / (1 +3.0563597)

S(3) = 72 / 4.0563597

S(3) = 17.749905 = $17.750 thousands

Amount of time required for sales to reach 70000

S(t) = 72/(1 + 9e^-0.36t)

S(t) = 70

70 = 72/(1 + 9e^-0.36t)

70 * (1 + 9e^-0.36t) = 72

(1 + 9e^-0.36t) = 72 / 70

1 + 9e^-0.36t = 1.0285714

9e^-0.36t = 1.0285714 - 1

9e^-0.36t = 0.0285714

e^-0.36t = 0.0285714 / 9

e^-0.36t = 0.0031746

Take the In of both sides ;

In(e^-0.36t) = In(0.0031746)

-0.36t = - 5.752573

t = - 5.752573 / - 0.36

t = 15.979

About 16 weeks

The limiting value in sales :

Take the limit as t - - > ∞

S(t - - > ∞) = 72/(1 + 9e^-0.36t)

Put t = 0

S(0) - - > 72 / (1 + 0)

72 / 1

= 72

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Since there are 12 buses and each one of them transports 40 students that means that the total amount of students will be equal to...

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a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-
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Answer:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

M = \frac{1}{2}(Lower + Upper)

Where

Lower \to Lower class interval

Upper \to Upper class interval

So, we have:

Class 63-65:

M = \frac{1}{2}(63 + 65) = 64

Class 66 - 68:

M = \frac{1}{2}(66 + 68) = 67

When the computation is completed, the frequency distribution will be:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

\bar x = 70.2903

\sigma = 3.5795

<em>See attachment for result of 1-VarStats</em>

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