NEED HELP ASAP!! Find the mode of the following data set.

Solve, graph, and give interval notation for the compound inequality:

Likurg_2 [28]

Answer:

The answer to your question is below

Step-by-step explanation:

Inequality 1

-4x + 1 > 13

-4x > 13 - 1

-4x > 12

x < 12/-4

x < -3

Inequality 2

4(x + 2) ≤ 4

4x + 8 ≤ 4

4x ≤ 4 - 8

4x ≤ - 4

x ≤ -4/4

x ≤ -1

Interval notation (-∞, -3)

See the graph below

The solution to these inequalities in where both intervals crosses

5 0

3 years ago

48% of men consider themselves professional baseball fans. you randomly select 10 men and ask each if he considers himself a pr

jekas [21]

If 48% consider themselves professional fans, you'll just have to apply this percentage to the number of selected men to find the answer:

(48 * 10) / 100 = number of professional baseball fans.

480 / 100 = number of professional baseball fans.

4.8

4.8 is not exactly five (a = incorrect)

4.8 is at least 6 (b = correct)

4.8 is not less than four (c = incorrect)

Hope it helped,

BioTeacher101

4 0

4 years ago

A brine solution of salt flows at a constant rate of 8L/min into a large tank that initially held 100L of brine solution in whic

GaryK [48]

Answer:

The mass of salt in the tank after t minutes is

$y(t) = 5-4.5e^{-\frac{2t}{25}}$

The concentration of salt in the tank reach 0.02 kg/L when $t=-\frac{25\ln \left(\frac{83}{75}\right)}{2} \approx -1.2669$

Step-by-step explanation:

Let y(t) be the mass of salt (in kg) that is in the tank at any time, t (in minutes).

The main equation that we will be using to model this mixing process is:

Rate of change of $\frac{dy}{dt}$ = Rate of salt in - Rate of salt out

We need to determine the rate at which salts enters the tank. From the information given we know:

The brine flows into the tank at a rate of $8\:\frac{L}{min}$
The concentration of salt in the brine entering the tank is $0.05\:\frac{kg}{L}$

The Rate of salt in = (flow rate of liquid entering) x (concentration of salt in liquid entering)

$(8\:\frac{L}{min}) \cdot (0.05\:\frac{kg}{L})=0.4 \:\frac{kg}{min}$

Next, we need to determine the output rate of salt from the tank.

The Rate of salt out = (flow rate of liquid exiting) x (concentration of salt in liquid exiting)

The concentration of salt in any part of the tank at time t is just y(t) divided by the volume. From the information given we know:

The tank initially contains 100 L and the rate of flow into the tank is the same as the rate of flow out.

$(8\:\frac{L}{min}) \cdot (\frac{y(t)}{100} \:\frac{kg}{L})= \frac{2y(t)}{25} \:\frac{kg}{min}$

At time t = 0, there is 0.5 kg of salt, so the initial condition is y(0) = 0.5. And the mathematical model for the mixing process is

$\frac{dy}{dt}=0.4-\frac{2y(t)}{25}, \quad{y(0)=0.5}$

$\frac{dy}{dt}=0.4-\frac{2y(t)}{25}\\\\\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}\\\\ \frac{1}{2}\frac{dy}{5-y}=\frac{1}{25}dt \\\\\frac{1}{2}\int \frac{dy}{5-y}=\int \frac{1}{25}dt\\\\\frac{1}{2}\left(-\ln \left|5-y\right|+C\right)=\frac{1}{25}t+C\\\\-\frac{1}{2}\ln \left|5-y\right|+C_1\right)=\frac{1}{25}t+C\\\\-\frac{1}{2}\ln \left|5-y\right|=\frac{1}{25}t+C_2\\\\5-y=C_3e^{-\frac{2t}{25} }\\\\y(t) =5-C_3e^{-\frac{2t}{25} }$

Using the initial condition y(0)=0.5

$y(t) =5-C_3e^{-\frac{2t}{25} }\\y(0)=0.5=5-C_3e^{-\frac{2(0)}{25}} \\C_3=4.5$

The mass of salt in the tank after t minutes is

$y(t) = 5-4.5e^{-\frac{2t}{25}}$

To determine when the concentration of salt is 0.02 kg/L, we solve for t

$y(t) = 5-4.5e^{-\frac{2t}{25}}\\\\0.02=5-4.5e^{-\frac{2t}{25}}\\\\5\cdot \:100-4.5e^{-\frac{2t}{25}}\cdot \:100=0.02\cdot \:100\\\\500-450e^{-\frac{2t}{25}}=2\\\\500-450e^{-\frac{2t}{25}}-500=2-500\\\\-450e^{-\frac{2t}{25}}=-498\\\\e^{-\frac{2t}{25}}=\frac{83}{75}\\\\\ln \left(e^{-\frac{2t}{25}}\right)=\ln \left(\frac{83}{75}\right)\\\\\frac{2t}{25}=\ln \left(\frac{83}{75}\right)\\\\t=-\frac{25\ln \left(\frac{83}{75}\right)}{2} \approx -1.2669$