(3 + 4i) + (5 − 2i) (2 points) −2 + 6i 2 − 2i 7 + 3i 8 + 2i

Reptile [31]

Answer:

Dion needs to add 10% of the solution

4 0

3 years ago

Colin invests £1200 into his bank account. He receives 3% per year simple interest. How much will Colin have after 4 years? Give

PolarNik [594]

Answer:

1350.61

Step-by-step explanation:

f(x)=1200(1.03^x)

3 0

3 years ago

In 2006,the General Social Survey (which is conducted uses a method similar to simple random sampling) included a question that

mars1129 [50]

Answer:

Assumptions are not met. Can not make confidence interval.

Step-by-step explanation:

In the General Social Survey, sample size is 1514.

The proportion of those who see themselves social is $\frac{470}{1514}$ ≈ 0.31

To give an 95% confidence interval, we should be able to calculate margin of error of the sample mean, which is given by the formula

M± $z*\frac{s}{\sqrt{N} }$ where M is the mean of the sample (in the General Social Survey it is 0.31), z is z-score for the 95% confidence level(approx. 1.94), s is the standard deviation of the sample, N is the size of the sample(in this example it is 1514).

Since we don't know the standard deviation of the sample, we cannot give a confidence interval.

8 0

3 years ago

You fly in a helicopter for 20 minutes that goes at a speed of 240 kilometers per hour. How far did you go?

wariber [46]

You have traveled 80km

3 0

3 years ago

Read 2 more answers

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

(c) Concentration of chlorine in the pool at time t

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

3 0

3 years ago