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3 years ago

5

the college textbook this year was 197. the following year it was 222. what is the percent change in the price of the book HEEEE

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1 answer:

romanna [79]3 years ago

3 0

197 divided 222 = 0.88% = 88%

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I'd really appreciate your help, please no joke answers

Lerok [7]

Answer:

d= 2√10

Step-by-step explanation:

y= 3x - 5

3x - y - 5=0

d= | 3* 7 +(-1) * (-4) + (+5) |

________________

$\sqrt{3^{2} } + (-1)^{2}$

d = | 20 | / $\sqrt{10}$

d = 20 / $\sqrt{10}$

d = 2 $\sqrt{10}$

3 0

3 years ago

A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min

Dmitry [639]

At any time $t$ (min), the volume of solution in the tank is

$500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}$

If $A(t)$ is the amount of salt in the tank at any time $t$ , then the solution has a concentration of $\dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}$ .

The net rate of change of the amount of salt in the solution, $A'(t)$ , is the difference between the amount flowing in and the amount getting pumped out:

$A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)$

Dropping the units and simplifying, we get the linear ODE

$A'=10+5\sin\dfrac t4-\dfrac A{10}$

$10A'+A=100+50\sin\dfrac t4$

Multiplying both sides by $e^{10t}$ allows us to identify the left side as a derivative of a product:

$10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}$

$\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}$

$e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt$

Integrate and divide both sides by $e^{10t}$ to get

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}$

The tanks starts off with 30 lb of salt, so $A(0)=30$ and we can solve for $C$ to get a particular solution of

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}$

6 0

3 years ago

A life insurance company sells a $100,000 one year term life insurance policy to a 30-year old male for $475. The probability th

laila [671]

Answer:

The expected value for the insurance company is $392.20.

Step-by-step explanation:

The expected value of a random variable, <em>X</em> is:

$E(X)=x\cdot P(X)$

It is provided that a life insurance company sells a $100,000 one year term life insurance policy to a 30-year old male for $475.

The probability that the male survives the year is, P(S) = 0.999172.

Then the probability that the male does not survives the year is:

P (S') = 1 - P (S)

= 1 - 0.999172

P (S') = 0.000828

The amount the company owes the male if he survives is, S = $475.

The amount the company owes the male if he does not survives is,

S' = $475 - $100,000 = -$99525.

Compute the expected value for the insurance company as follows:

$E(\text{Insurance Company})=S\cdot P(S)+S'\cdot P(S')$

$=(475\times 0.999172)+(-99525\times 0.000828)\\=474.6067-82.4067\\=392.20$

Thus, the expected value for the insurance company is $392.20.

4 0

3 years ago

I will give you the crown if you answer this.

babymother [125]

Answer:

Number 6 is -0.8571

Number 7 is 0.3

Step-by-step explanation:

8 0

2 years ago

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A coffee canister holds 12 ounces of coffee. A coffee scoop holds 3/4 of an ounce of coffee. How many scoops of coffee are in th

arsen [322]

Answer: D

Step-by-step explanation:

3/4= 0.75

0.75÷12=16 scoops

8 0

3 years ago

Read 2 more answers