<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B3%7D%20%20-%20%20%5Cfrac%7B4%7D%7B10%7D%20%20%3D%20" id="TexFormula1" titl

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3 years ago

e=" \frac{2}{3} - \frac{4}{10} = " alt=" \frac{2}{3} - \frac{4}{10} = " align="absmiddle" class="latex-formula">
what is the answer?

Mathematics

1 answer:

rusak2 [61]3 years ago

6 0

Answer:

8/30 (Simplified answer: 4/15)

Step-by-step explanation:

Find common denominators. Note the what you do to the denominator, you do to the numerator.

Multiply 10 to both the denominator & numerator of 2/3, and multiply 3 to both the denominator & numerator of 4/10

(10/10)(2/3) - (3/3)(4/10)

Simplify.

(20/30) - (12/30)

Simplify. Combine the terms

(20/30) - (12/30) = (20 - 12)/30 = 8/30

Simplify the fraction

(8/30)/(2/2) = 4/15

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Find the value of x. The dot represents the center of the circle,

djyliett [7]

Answer:

$x=16\ units$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

Triangle ABC and ADE are congruent triangles by SSS Theorem postulate

therefore

$x=8(2)=16\ units$

5 0

3 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}}$

3 0

3 years ago

What is the probability it will be a sunny day a year from today?

s344n2d4d5 [400]

Assuming that your choices are sunny and not sunny. There would be a 50% chance that a year from today will be sunny.

Therefore the answer would be 1/2 or 50%.

5 0

3 years ago

Read 2 more answers

Which of the following numbers are greater than or equal to 4/7 A. 2/5 B. 0.57 C. 2/3

Bumek [7]

Answer:

Step-by-step explanation:

If you were to divide 4 by 7, you would get a decimal that would be rounded up to that value. 2/3 is roughly .66 and 2/5 is .4.

6 0

3 years ago

In a recent year, an author wrote 169 checks. Use the Poisson distribution to find the probability that, on a randomly selecte

grigory [225]

We should first calculate the average number of checks he wrote per day. To do that, divide 169 by 365 (the number of days in a year) and you get (rounded) 0.463. This will be λ in our Poisson distribution. Our formula is
$P(X=k)= \frac{ \lambda ^{k}-e^{-\lambda} }{k!}$ . We want to evaluate this formula for X≥1, so first we must evaluate our case at k=0.
$P(X=0)= \frac{0.463 ^{0}-e ^{-0.463} }{0!} \\ = \frac{1-e ^{-0.463} }{1} =0.3706$
To find P(X≥1), we find 1-P(X<1). Since the author cannot write a negative number of checks, this means we are finding 1-P(X=0). Therefore we have 1-0.3706=0.6294.
There is a 63% chance that the author will write a check on any given day in the year.

8 0

3 years ago