PLEASE HELP! much appreciated :D<br><br> Find the value of x.

Kisachek [45]

28/36 will be your answer which can be simplified into 14/18 which can be simplified again to 7/9 and that's your final answer

7 0

4 years ago

I need help with #15

gladu [14]

15. MP(5,11) S(3,5) M(x,y)

(3+x)/2 = 5 x=7

(5+y)/2 =-11 y=-27

M(7,-27) is the other point

8 0

4 years ago

There are a total of 127 cars and trucks in a lot. If there are four more than twice the number of tricks than cars, how many ca

Kruka [31]

There are a total of 127 cars on a lot if there are four more than twice the number of trucks than cars how any trucks and cars are on the lot

4 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

Suppose the weights of farmer carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1

denis23 [38]

Given that <span>the weights of farmer carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1 ounces.

The probability of a normally distributed data between two values (a, b) is given by:

$P(a\ \textless \ X\ \textless \ b)=P\left(z\ \textless \ \frac{b-\mu}{\sigma/\sqrt{n}} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma/\sqrt{n}} \right) \\ \\ =P(7.5\ \textless \ X\ \textless \ 8.5)=P\left(z\ \textless \ \frac{8.5-8.0}{1.1/\sqrt{6}} \right)-P\left(z\ \textless \ \frac{7.5-8.0}{1.1/\sqrt{6}} \right) \\ \\ =P(z\ \textless \ 1.113)-P(-1.113)=0.8672-0.1328=0.7344$ </span>

6 0

4 years ago

PLEASE HELP! much appreciated :D Find the value of x.