What comes after 93 27 14.

NEED HELP ASAP!!! ON A KHAN ACADEMY TEST AND DON"T HAVE MUCH TIME LEFT!!! CORRECT ANSWER GETS BRAINLIEST!!!

Otrada [13]

Answer:

Step-by-step explanation:

This is correct, except that if you have two variables you will need two equations. The other equation (the easy one!) was x+ y = 100, from which you might want to solve for y = 100-x, and substitute into the other equation to get:

.30x + .02(100-x) = .04(100)

.30x + 2 - .02x = 4

.28x = 2

x = 2/.28 = 50/7 or approximately 7.14 gallons of cream

100-x = 92 6/7 or approximately 92.86 gallons of 2%

R^2 at SCC

6 0

1 year ago

A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o

Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

First, we determine the Rate In

Rate In=(concentration of salt in inflow)(input rate of brine)

$=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}$

Change In Volume of the tank, $\frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}$

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

Rate Out

Rate Out=(concentration of salt in outflow)(output rate of brine)

$R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}$

Therefore:

$\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3$

This is a linear differential equation in standard form, therefore the integrating factor:

$e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2$

Multiplying the DE by the integrating factor, we have:

$(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}$

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

$20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000$

Therefore, the amount of salt in the tank at any time t is:

$A(t)=(50+t)-75000(50+t)^{-2}$

After 15 minutes, the amount of salt in the tank is:

$A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds$

8 0

1 year ago

Javier writes an integer on a piece of paper. The absolute value of his number is less than 4. Which is true of the possible val

ANTONII [103]

The possible values are less than 4 but greater than –4.

7 0

1 year ago

Read 2 more answers

The means and mean absolute deviations of the amount of rain that fell each day in a local city, last week and this week, are sh

Dennis_Churaev [7]

The difference in the mean rainfall is given by 19.7 - 19.3 = 0.5

The difference in the mean rainfall is approximately 0.5/4.6 = 0.087 = 8.7 percent of the mean absolute deviation for last week and is 0.5/5.2 = 0.077 = 7.7 percent of the mean absolute deviation for this week.

Therefore, the difference in the mean rainfall is approximately what percent of the mean absolute deviation 8% of the mean absolute deviation of the data sets.

4 0

1 year ago

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One division of a large defense contractor manufactures telecommunication equipment for the military. This division reports that

Zielflug [23.3K]

Answer:

At the 95% confidence level, management should not conclude that the percentage of rework for electrical components is lower than the rate of 12% for non-electrical components, since the upper bound of the confidene interval, which is 0.1339, is higher than 0.12.

Step-by-step explanation:

To reach a conclusion, we have to observe the confidence interval.

Should management conclude that the percentage of rework for electrical components is lower than the rate of 12% for non-electrical components?

Is the upper bound of the confidence interval lower than 12% = 0.12?

If yes, it should conclude that this percentage is lower.

Otherwise, it cannot conclude.

Confidence interval:

0.0758 to 0.1339

0.1339 is higher than 0.12.

So.

At the 95% confidence level, management should not conclude that the percentage of rework for electrical components is lower than the rate of 12% for non-electrical components, since the upper bound of the confidene interval, which is 0.1339, is higher than 0.12.

6 0

1 year ago

Read 2 more answers