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Archy [21]
3 years ago
7

Two cars, car X and car Y , start moving from the same point P on a cross intersection. Car X is travelling east and car Y is tr

avelling north. Some time later car X is 60 km east of point P and travelling in an easterly direction at 80 km/h and car Y is 80 km north of point P and travelling in a northerly direction at 100 km/h. How fast is the distance between car X and car Y changing?
Mathematics
1 answer:
agasfer [191]3 years ago
5 0
Let the distance traveled by car X be x km 
<span>let the distance traveled by car Y by y km </span>
<span>their paths form a right-angled triangle. </span>
<span>Let the distance between them be D km </span>
<span>D^2 = x^2 + y^2 </span>
<span>2D dD/dt = 2x dx/dt + 2y dy/dt </span>
<span>dD/dt = (x dx/dt + y dy/dt)/D </span>

<span>at the given case: </span>
<span>x = 60, y = 80 , dx/dt = 80, dy/dt = 100 </span>
<span>D^2 = 60^2 + 80^2 = 10000 </span>
<span>D = 100 </span>

<span>dD/dt = (60(80) + 80(100))/100 </span>
<span>= 128 </span>
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Two trains leave towns 664 miles apart at the same time and travel toward each other. One train travels 16 mih faster than the o
Hatshy [7]

Answer:

I think that I am right.  One train is traveling at 75 mph and the other train is traveling at 91 mph

Step-by-step explanation:

5 0
1 year ago
A stereo store is offering a special price on a complete set of components (receiver, compact disc player, speakers, turntable).
Juli2301 [7.4K]

Answer:

a) 240 ways

b) 12 ways

c) 108 ways

d) 132 ways

e) i) 0.55

ii) 0.4125

Step-by-step explanation:

Given the components:

Receiver, compound disk player, speakers, turntable.

Then a purcahser is offered a choice of manufacturer for each component:

Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood => 5 offers

Compact disc player: Onkyo, Pioneer, Sony, Technics => 4 offers

Speakers: Boston, Infinity, Polk => 3 offers

Turntable: Onkyo, Sony, Teac, Technics => 4 offers

a) The number of ways one component of each type can be selected =

\left(\begin{array}{ccc}5\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right)

= 5 * 4 * 3 * 4  = 240 ways

b) If both the receiver and compact disk are to be sony.

In the receiver, the purchaser was offered 1 Sony, also in the CD(compact disk) player the purchaser was offered 1 Sony.

Thus, the number of ways components can be selected if both receiver and player are to be Sony is:

\left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right)

= 1 * 1 * 3 * 4 = 12 ways

c) If none is to be Sony.

Let's exclude Sony from each component.

Receiver has 1 sony = 5 - 1 = 4

CD player has 1 Sony = 4 - 1 = 3

Speakers had 0 sony = 3 - 0 = 3

Turntable has 1 sony = 4 - 1 = 3

Therefore, the number of ways can be selected if none is to be sony:

\left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right)

= 4 * 3 * 3 * 3 = 108 ways

d) If at least one sony is to be included.

Number of ways can a selection be made if at least one Sony component is to be included =

Total possible selections - possible selections without Sony

= 240 - 108

= 132 ways

e) If someone flips switches on the selection in a completely random fashion.

i) Probability of selecting at least one Sony component=

Possible selections with at least one sony / Total number of possible selections

\frac{132}{240} = 0.55

ii) Probability of selecting exactly one sony component =

Possible selections with exactly one sony / Total number of possible selections.

\frac{\left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) + \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) + \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}1\\1\end{array}\right)}{240}

= \frac{(1*3*3*3)+(4*1*3*3)+(4*3*3*1)}{240}

\frac{27 + 36 + 36}{240} = \frac{99}{240} = 0.4125

5 0
3 years ago
Volume ....................
Ilia_Sergeevich [38]

Answer:

360 cm^2

Step-by-step explanation:

length x width x height

6 x 6 x 10

3 0
3 years ago
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How many ways are there to distribute five balls into seven boxes if?
katovenus [111]
There are a lot of ways but here are a couple of examples.
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3 0
3 years ago
A score that is three standard deviations above the mean would have a z score of
tia_tia [17]

The value of z-score for a score that is three standard deviations above the mean is 3.

In this question,

A z-score measures exactly how many standard deviations a data point is above or below the mean. It allows us to calculate the probability of a score occurring within our normal distribution and enables us to compare two scores that are from different normal distributions.

Let x be the score

let μ be the mean and

let σ be the standard deviations

Now, x =  μ + 3σ

The formula of z-score is

z_{score} = \frac{x-\mu}{\sigma}

⇒ z_{score} = \frac{\mu + 3\sigma -\mu}{\sigma}

⇒ z_{score} = \frac{ 3\sigma }{\sigma}

⇒ z_{score} = 3

Hence we can conclude that the value of z-score for a score that is three standard deviations above the mean is 3.

Learn more about z-score here

brainly.com/question/13448290

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6 0
2 years ago
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