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Sauron [17]
1 year ago
13

A small airplane flies 1200 miles with an average speed of 300 miles per hour. 2 hours after the plane leaves, a Boeing 747 leav

es from the same point. Both planes arrive at the same time.what was the average speed of the boeing 747?
Mathematics
1 answer:
inysia [295]1 year ago
8 0

The small plane has the following parameters

Distance = 1200 miles

Speed = 300miles/hour

time = ?

\begin{gathered} \text{Speed =}\frac{dis\tan ce}{time} \\ time\text{= }\frac{1200}{300} \\ \text{time = 4 hours} \end{gathered}

If a Boeing 747 leaves 2 hours later, the total time becomes 4hours + 2 hours = 6 hours

Considering the distance of the small plane after 6 hours, we have

\begin{gathered} \text{distance = sp}eed\times\text{ time} \\ d\text{ = 300 }\times6 \\ d\text{ =1800 miles} \end{gathered}

So the total distance = 1800 miles

Hence the speed of the Boeing 747 will be

\begin{gathered} \text{Speed =}\frac{dis\tan ce}{time} \\ \text{distance = 1800 miles } \\ time\text{ = 2hours} \end{gathered}\begin{gathered} \text{speed = }\frac{1800}{2} \\ \text{speed = 900 miles/hour} \end{gathered}

The average speed of the Boeing 747 is 900 miles/ hour

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3 years ago
Weights and heights of turkeys tend to be correlated. For a population of turkeys at a farm, this correlation is found to be 0.6
LenaWriter [7]

Answer:

a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than <u>79.37 %</u> of them.

The  average height for turkeys at the 90th percentile for weight is 34.554

Of the turkeys at the 90th percentile for weight, roughly the percentage that  would  be taller than 28 inches 79.37%

Step-by-step explanation:

Given that:

For a population of turkeys at a farm, the correlation found between the weights and heights of turkeys is r = 0.64

the average weight in pounds \overline x = 17

the standard deviation of the weight in pounds S_x = 5

the average height in inches \overline y = 28

the standard deviation of the height in inches S_y = 8

Also, given that the weight and height both roughly follow the normal curve

For this study , the slope of the regression line can be expressed as :

\beta_1 = r \times ( \dfrac{S_y}{S_x})

\beta_1 = 0.64 \times ( \dfrac{8}{5})

\beta_1 = 0.64 \times 1.6

\beta_1 = 1.024

To the intercept of the regression line, we have the following equation

\beta_o = \overline y - \beta_1 \overline x

replacing the values:

\beta_o = 28 -(1.024)(17)

\beta_o = 28 -17.408

\beta_o = 10.592

However, the regression line needed for this study can be computed as:

\hat Y = \beta_o + \beta_1 X

\hat Y = 10.592 + 1.024 X

Recall that;

both the weight and height roughly follow the normal curve

As such, the weight related to 90th percentile can be determined as shown below.

Using the Excel Function at 90th percentile, which can be computed as:

(=Normsinv (0.90) ; we have the desired value of 1.28

∴

\dfrac{X - \overline x}{s_x } = 1.28

\dfrac{X - 17}{5} = 1.28

X - 17 = 6.4

X = 6.4 + 17

X = 23.4

The predicted height \hat Y = 10.592 + 1.024 X

where; X = 23.4

\hat Y = 10.592 + 1.024 (23.4)

\hat Y = 10.592 + 23.9616

\hat Y = 34.5536

Now; the probability of predicted height less than 34.5536 can be computed as:

P(Y < 34.5536) = P( \dfrac{Y - \overline y }{S_y} < \dfrac{34.5536-28}{8})

P(Y < 34.5536) = P(Z< \dfrac{6.5536}{8})

P(Y < 34.5536) = P(Z< 0.8192)

From the Z tables;

P(Y < 34.5536) =0.7937

Hence,  a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than <u>79.37 %</u> of them.

The  average height for turkeys at the 90th percentile for weight is :

\hat Y = 10.592 + 1.024 X

where; X = 23.4

\hat Y = 10.592 + 1.024 (23.4)

\hat Y = 10.592 + 23.962

\mathbf{\hat Y = 34.554}

Of the turkeys at the 90th percentile for weight, roughly what percent would you estimate to be taller than 28 inches?

i.e

P(Y >28) = 1 - P (Y< 28)

P(Y >28) = 1 - P( Z < \dfrac{28 - 34.554}{8})

P(Y >28) = 1 - P( Z < \dfrac{-6.554}{8})

P(Y >28) = 1 - P( Z < -0.8193)

From the Z tables,

P(Y >28) = 1 - 0.2063

\mathbf{P(Y >28) = 0.7937}

= 79.37%

7 0
4 years ago
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