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Arada [10]
3 years ago
8

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,<br>​
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}

(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
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<br> A. 2/5<br> B. 0.57<br> C. 2/3
Bumek [7]

Answer:

B.

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.<em />
8 0
3 years ago
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