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3 years ago

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A bus can hold a maximum of 68 passengers. There are 14 passengers on the bus. The same number of passengers get on the bus at e

ach of the next 6 stops. Write an inequality to represent the number of passengers, p, that can get on the bus at each of the stops.

1 answer:

mr_godi [17]3 years ago

7 0

Answer:

68p + 14 = 6

Step-by-step explanation:

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The people with smartphones is 2000 and the people with flipphones is 1500. the ratio is 2000:1500

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A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito th

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Answer:

Part a)

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b)

Check the attached figure to see the ultimate behavior of the graph.

Part c)

The level = 25, Amplitude = 0.2499

Step-by-step Solution:

Part a)

Given:

$Q(0)=50$

Rate in:

$\frac{1}{4}\left(1+\frac{1}{2}sint\right)\cdot 2\:=\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

Rate out:

$\frac{Q}{100}\cdot 2=\frac{Q}{50}$

So, the differential equation would become:

$\frac{dQ}{dt}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)-\frac{Q}{50}$

Rewriting the equation:

$\frac{dQ}{dt}+\frac{Q}{50}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

As $p(x)$ is the coefficient of $y$ , while $q(x)$ is the constant term in the right side of the equation:

$p\left(x\right)=\frac{1}{50}$

$q\left(x\right)=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

First it is important to determine the function $\mu$ :

$\mu \left(t\right)=e^{\int \:p\left(t\right)dt}$

$=e^{\int \:\left(\frac{1}{50}\right)dt}$

$=e^{\frac{t}{50}}$

The general solution then would become:

$y\left(t\right)=\frac{1}{\mu \left(t\right)}\left(\int \mu \left(t\right)q\left(t\right)dt+c\:\right)$

$=\frac{1}{e^{\frac{t}{50}}}\int e^{\frac{t}{50}}\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)dt+\frac{1}{e^{\frac{t}{50}}}c$

$=\frac{1}{e^{\frac{t}{50}}}\left(\frac{-25e^{\frac{t}{50}}\left(50cost-sint\right)}{5002}+25e^{\frac{t}{50}}\right)+\frac{1}{e^{\frac{t}{50}}}c$

$=\frac{\left-1250cost+25sint\right}{5002}+25+\frac{1}{e^{\frac{t}{50}}}c$

Evaluate at $t=0$

$50=y\left(0\right)=\frac{\left(-1250cos0+25sin0\right)}{5002}+25+\frac{1}{e^{\frac{0}{50}}}c$

Solve to c:

$c=25+\frac{1250}{5002}$

$\mathrm{Cancel\:}\frac{1250}{5002}:\quad \frac{625}{2501}$

$c=25+\frac{625}{2501}$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:25=\frac{25\cdot \:2501}{2501}$

$c=\frac{25\cdot \:2501}{2501}+\frac{625}{2501}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$c=\frac{25\cdot \:2501+625}{2501}$

$c=\frac{63150}{2501}$

$c\approx 25.25$

Therefore, the general solution then would become:

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b) <em>Plot the Solution to see the ultimate behavior of the graph</em>

The graph appears to level off at about the value of $Q=25$ .

The graph is attached below.

Part c)

In the graph we note that the level is $Q=25$ .

Therefore, the level = 25

The amplitude is the (absolute value of the) coefficient of $cost\:t$ in the general solution (as the coefficient of the sine part is a lot smaller):

Therefore,

$A=\frac{1250}{5002}\:\approx 2.499$

Keywords: differential equation, word problem

Learn more about differential equation word problem from brainly.com/question/14614696

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Answer:

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Answer: MAD= 3

Step-by-step explanation:

Find the mean of the data set

31+24+32+28+29+36/6= 30

subtract EACH data by the mean, find the absolute value of the answer

31-30= 1 (absolute value)

24-30= 6 (absolute value)

32-30= 2 (absolute value)

28-30= 2 (absolute value)

29-30= 1 (absolute value)

36-30= 6 (absolute value)

Now find the mean of the values

1+6+2+2+1+6/6=3

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4 years ago

Last Month, Levi had written 125 words of his book. This month,he is up to 12,580 words. About how many times more words has he

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Answer:

He has written about 100 times as many words.

Step-by-step explanation:

In order to find this, we first need to find the number that he has written this month. In order to do so, we need to subtract the previous month from the current total.

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Now that we have that number, we can divide it by the previous month to see how many times more words he wrote in the second month.

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