In January, Wally's Widget World had sales of $12,500 and expenses of $10,200. The profit for January was ____

At a college the scores on the chemistry final exam are approximately normally distributed, with a mean of 75 and a standard dev

AysviL [449]

Answer:

b) Chemistry

Step-by-step explanation:

To compare both scored we need to standardize the scores using the following equation:

$\frac{x-m}{s}$

Where x is the score, m is the mean and s is the standard deviation. So, 82 on chemistry is equivalent to:

$\frac{82-75}{15}=0.4667$

Because the mean of the scores on the chemistry final exam is equal to 75 and the standard deviation is 15

At the same way, 80 on Calculus is equivalent to:

$\frac{80-83}{13} =-0.2308$

Because the mean of the scores on the calculus final exam is equal to 83 and the standard deviation is 13

Now, we can compare the values. So, taking into account that -0.2308 is lower than 0.4667, we can said that the student do better in Chemistry.

5 0

3 years ago

A painter spends 3 hours working on a painting. A sculptor spends 2 and 2/3 times as long working on a sculpture. How long does

Anon25 [30]

Answer:

8 hours

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows