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3 years ago

BRAINLIEST URGENTTT

Mathematics

2 answers:

marusya05 [52]3 years ago

8 0

Answer:

$890,600

Step-by-step explanation:

alex41 [277]3 years ago

4 0

Answer:

They say you should save 10%

So if you calculate her yearly income

3420 x 26 pay periods per year = 88,920

Multiply it by 10

$889,200

Step-by-step explanation:

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Gabbys age is two times mikhails age.the sum of their ages is 75. What is mikhails age

Tatiana [17]

Answer:

Mikhail's age is 25 years old

Step-by-step explanation:

Let's make an equation: Say Mikhail is x, so Gabby would be 2*x=2x

If the sum of their ages equals 75, then x+2x=75, and x=2x=3x, so 3x=75.

75/3=25

6 0

3 years ago

7. Give the x-intercept and y-intercept of the graph and three more solutions

Sveta_85 [38]

Answer:

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2 years ago

Suppose a researcher is interested in understanding the variation in the price of store brand milk. A random sample of 36 grocer

liberstina [14]

Answer: ($3.055, $3.205)

Step-by-step explanation:

Given : Significance level : $\alpha: 1-0.95=0.5$

Critical value : $z_{\alpha/2}=1.96$

Sample size : n= 36

Sample mean : $\overline{x}=\$\ 3.13$

Standard deviation : $\sigma= \$\ 0.23$

The confidence interval for population mean is given by :_

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

$\text{i.e. }\$\ 3.13\pm (1.96)\dfrac{0.23}{\sqrt{36}}\\\\\approx\$\ 3.13\pm0.075\\\\=(\$\ 3.13-0.075,\$\ 3.13+0.075)=(\$\ 3.055,\$\ 3.205)$

Hence, the 95% confidence interval to estimate the population mean = ($3.055, $3.205)

3 0

2 years ago

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)

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

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}}$

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3 years ago

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GuDViN [60]

Answer:

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Step-by-step explanation:

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3 years ago