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1 year ago

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A researcher wishes her patients to try a new medicine for depression. How many different ways can she select 5 patients from 50

patients?

1 answer:

Readme [11.4K]1 year ago

7 0

Answer:

she can select 10 different ways

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Lenovo uses the zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as follows:

Stella [2.4K]

Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

$\begin{tabular}
{|c|c|c|c|}
Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
\end{tabular}$

The forcast for a period $F_{t+1}$ is given by the formular

$F_{t+1}=\alpha A_t+(1-\alpha)F_t$

where $A_t$ is the actual value for the preceding period and $F_t$ is the forcast for the preceding period.

Part 1A:
Given <span>α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.1(1.57)+(1-0.1)(1.83) \\ \\ =0.157+0.9(1.83)=0.157+1.647 \\ \\ =1.804$

Therefore, the foreast for period 11 is $1.80

Part 1B:

</span>Given <span>α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ 
=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

Therefore, the foreast for period 12 is $1.78</span>

Part 2A:

Given <span>α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ 
=0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703$

Therefore, the foreast for period 11 is $1.70

</span>
<span><span>Part 2B:

</span>Given <span>α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ 
=0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676$

Therefore, the foreast for period 12 is $1.68

</span></span>
<span>Part 3A:

Given <span>α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ 
=0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645$

Therefore, the forecast for period 11 is $1.65

</span>
<span><span>Part 3B:

</span>Given <span>α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ 
=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

Therefore, the forecast for period 12 is $1.64

Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span></span></span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:

$\frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3} \\ \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157

</span><span><span>Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = 
\frac{|-0.17|+|-0.08|+|-0.07|}{3} \\ \\ = \frac{0.17+0.08+0.07}{3} = 
\frac{0.32}{3} \approx0.107$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107

</span></span>
<span><span>Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span><span>α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = 
\frac{|-0.15|+|-0.03|+|0.11|}{3} \\ \\ = \frac{0.15+0.03+0.11}{3} = 
\frac{29}{3} \approx0.097$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>

5 0

1 year ago

Given that r||s and q is a transversal, we know that by the [________]. corresponding angles theorem alternate interior angles t

neonofarm [45]

Answer:

alternate interior angles theorem

Step-by-step explanation:

The alternate interior angles theorem states that when two parallel lines are cut by a transversal, the resulting angles produced are a pair of congruent alternate interior angles.

Given the image attached below, both line k and line l are parallel to each other and also, line t is the transversal, therefore the resulting congruent alternate interior angles produced are:

∠ 4 ≅ ∠6, ∠1 ≅ ∠7

5 0

2 years ago

Read 2 more answers

Which statements are true about the graph of y s 3x + 1 and y 2 -x + 2? Check all that apply.

Varvara68 [4.7K]

Let's go through all the options one at a time and look for the correct ones

<u>Option 1: The slope of one boundary line is 2</u>

We have 2 equations of lines, where the coefficients of are 3 and -1 respectively

because the coefficient of x denotes the slope of a line, we know that the lines have the slope 3 and -1, not 2

Hence, this option is Incorrect

<u>Option 2: Both boundary lines are solid</u>

In order for the boundary lines to be solid, the inequality must have an 'equal to', like ≤ (less than or equal to) or ≥ (greater than or equal to)

we can see that that's the case in our case and hence, this option is Correct

<u>Option 3: A solution to the system is (1, 3)</u>

To confirm this, we'll plug these coordinates into the given inequalities and see if it stands correct

y ≤ 3x + 1

3 ≤ 3(1) + 1

3 ≤ 4 which is correct because 3 is less than 4

Second equation:

y ≥ 2 - x

3 ≥ 2 - 1

3 ≥ 1

Which is also true because 3 is greater than 1

Now, we can say that (1 , 3) is a solution to the system because it satisfies both the equations and is Correct

<u>Option 4: Both inequalities are shaded below the boundary lines</u>

For an inequality to be shaded below the boundary line, it must have the ≤ inequality (in case of solid line) and < inequality (in case of dotted line)

because the second inequality listed includes the ≥ inequality, which was not mentioned above, it won't be shaded below

another way to think about it is that any 'greater than' inequality will shade everything above the line and the 'lesser than' inequality will shade below the line

which means that this option is Incorrect

<u>Option 5: The boundary lines intersect</u>

In order for the boundary lines to intersect, they must have have different slopes.

as we mentioned in the explanation of the first option, that the slopes of the lines is 3 and -1, which are different slopes

Therefore, this option is Correct

5 0

1 year ago

Read 2 more answers

The cost of a parking permit consists of a one-time administration fee plus a monthly fee. A permit purchased for 12 months cost

LekaFEV [45]

12x50=600 so 60 left
15x50=750 so 60 left
so the administration fee is 60

5 0

2 years ago

Read 2 more answers

If the instructions for a problem ask you to use the smallest possible domain to completely graph two periods of y = 5 + 3 cos 2

erica [24]

Hello,

y=5+3*cos (2(x-π/3))

The function is periodic with periode=2π.

-1<=cos (2(x-π/3))<=1
==>-1*3<=3*cos (2(x-π/3))<=3*1
==>5-3<=5+3cos(2(x-π/3))<=5+3
==>2<= y<=8

8 0

1 year ago