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kenny6666 [7]
3 years ago
12

At maximum speed, an airplane tracked 1720 miles against the wind in 5 hours. Flying with the wind, the plane can travel the sam

e distance in 4 hours.
Let x be the maximum speed of the plane and y be the speed of the wind. What is the speed of the plane with no wind?
Mathematics
1 answer:
nataly862011 [7]3 years ago
6 0
Recall your d = rt, distance = rate * time.

as the plane goes against the wind, the plane is not really flying at "x" mph, but is really going slower, at " x - y ", because the wind is subtracting speed from it.

likewise, when the plane is going with the wind, is not really going at "x" mph, but at " x + y ", is going faster due to the wind, thus

\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{against the wind}&1720&x-y&5\\
\textit{with the wind}&1720&x+y&4
\end{array}
\\\\\\
\begin{cases}
1720=(x-y)5\implies \frac{1720}{5}=x-y\\\\
344=x-y\implies \boxed{y}=x-344\\
---------------\\
1720=(x+y)4\implies \frac{1720}{4}=x+y\\\\
430=x+y\implies 430=x+\left( \boxed{x-344} \right)
\end{cases}
\\\\\\
430=2x-344\implies 774=2x\implies \cfrac{774}{2}=x\implies 387=x
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Step-by-step explanation:

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If a rectangle has a length of 8 centimeters and a width of 5 centimeters, then what is the area of the rectangle in square cent
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The lengths of lumber a machine cuts are normally distributed with a mean of 106 inches and a standard deviation of 0.3 inch. ​(
Vinvika [58]

Answer:

a) P(X>106.11)=P(\frac{X-\mu}{\sigma}\frac{106.11-106}{0.3})=P(z>0.37)

And we can find this probability with the complement rule:

P(z>0.37)=1-P(z

b) z =\frac{106.11- 106}{\frac{0.3}{\sqrt{44}}}=  2.431

And if we use the z score we got:

P(z>2.431) =1-P(z

Step-by-step explanation:

Let X the random variable that represent the lengths of a population, and for this case we know the distribution for X is given by:

X \sim N(106,0.3)  

Where \mu=106 and \sigma=0.3

Part a

We are interested on this probability

P(X>106.11)

And we can use the z score formula given by:

z=\frac{x-\mu}{\sigma}

And using this formula we got:

P(X>106.11)=P(\frac{X-\mu}{\sigma}\frac{106.11-106}{0.3})=P(z>0.37)

And we can find this probability with the complement rule:

P(z>0.37)=1-P(z

Part b

For this case we select a sample of n =44 and the new z score formula is given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score we got:

z =\frac{106.11- 106}{\frac{0.3}{\sqrt{44}}}=  2.431

And if we use the z score we got:

P(z>2.431) =1-P(z

6 0
3 years ago
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