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Varvara68 [4.7K]

3 years ago

11

The scores of eight grade students in a math test are normally distributed with a mean of 57.5 and a standard deviation of 6.5.

From this data, we can conclude that 68% of the students recieved scores between
a)34 and 64
b)34 and 68
c)51 and 64
d)51 and 68

1 answer:

rewona [7]3 years ago

8 0

Work out the z scores for 51 and 54
z1 = (51 - 57.5) / 6.5 = -1
z2 = (64 - 57.5)/6.5 = 1
from tables of normal distribution this value os 3413 + 3413 = 68.26%

so the answer is c

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liq [111]

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Step-by-step explanation:

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6 0

2 years ago

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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

4 0

3 years ago

Help me please please

Vera_Pavlovna [14]

Answer:

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Step-by-step explanation:

The angle at Port Latta between the actual direction of travel and the destination at Lookout Point is 90° -75° = 15°. The sine function relates the angle, the side opposite, and the hypotenuse of a right triangle, so you have ...

sin(15°) = (distance to Lookout Point)/(15 nm)

Solving for the desired distance, we have ...

distance to Lookout Point = (15 nm)sin(15°) = 3.88 nm

The yacht is 3.88 nautical miles from Lookout Point.

3 0

3 years ago

I have 3 more help please

ira [324]

Ok -13*-8 is 104.
-8*18q is 144q.
Combine an you get:
144q+104

3 0

2 years ago