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Liula [17]
2 years ago
14

Dalia flies an ultralight plane with a tailwind to a nearby town in 1/3 of an hour. On the return trip, she travels the same dis

tance in 3/5 of an hour. What is the average rate of speed of the wind and the average rate of speed of the plane?
Initial trip:
Return trip:
Let x be the average airspeed of the plane.
Let y be the average wind speed.
Initial trip: 18 = (x + y)
Return trip: 18 = (x – y)
Dalia had an average airspeed of miles per hour.
The average wind speed was miles per hour.
Mathematics
2 answers:
user100 [1]2 years ago
9 0

Answer:

the average airspeed of the plane = 42 miles per hour

the average wind speed = 12 miles per hour

Step-by-step explanation:

Let x be the average airspeed of the plane.

Let y be the average wind speed.

Distance =time * speed

Initial trip: 18 = \frac{1}{3}(x+y)

Return trip: 18 = \frac{3}{5}(x+y)

We solve for x  and y

18 = \frac{1}{3}(x+y)

Multiply both sides by 3

54= x+ y

y= 54- x ------------> equation 1

18 = \frac{3}{5}(x+y)

Multiply both side by 5

90 = 3(x-y)

90= 3x- 3y ------------------> equation 2

Plug in y=54-x in second equation

90= 3x- 3(54-x)

90 = 3x - 162 + 3x

90 = -162 + 6x

Add 162 on both sides

252= 6x

Divide both sides by 6

So x= 42

y= 54- x

Plug in 42 for x

y= 54 - 42= 12

the average airspeed of the plane = 42 miles per hour

the average wind speed = 12 miles per hour




pashok25 [27]2 years ago
8 0
Dalia had an average airspeed of ⇒ 42 miles per hour.
The average wind speed was  ⇒ 12 miles per hour.

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4. B. 1.059

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Step-by-step explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

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Part 1

The hypothesis for this case are:

Null hypothesis: \mu_{A}=\mu_{B}=\mu_{C}

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=A,B,C

Part 2

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have p groups and on each group from j=1,\dots,p we have n_j individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2

And we have this property

SST=SS_{between}+SS_{within}

We need to find the mean for each group first and the grand mean.

\bar X =\frac{\sum_{i=1}^n x_i}{n}

If we apply the before formula we can find the mean for each group

\bar X_A = 27, \bar X_B = 24, \bar X_C = 30. And the grand mean \bar X = 27

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SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2

Each group have a sample size of 4 so then n_j =4

SS_{between}=SS_{model}=4(27-27)^2 +4(24-27)^2 +4(30-27)^2=72

The degrees of freedom for the variation Between is given by df_{between}=k-1=3-1=2, Where  k the number of groups k=3.

Now we can find the mean square between treatments (MSTR) we just need to use this formula:

MSTR=\frac{SS_{between}}{k-1}=\frac{72}{2}=36

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Part 3

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C. 34

Part 4

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F=\frac{MSTR}{MSE}=\frac{36}{34}=1.059

B. 1.059

Part 5

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And we can use excel to find this critical value with this function:

"=F.INV(1-0.01,2,9)"

And we will see that the critical value is F_{crit}=8.02

B. 8.02

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