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DanielleElmas [232]
4 years ago
13

A factory has two bottle filling machines, which run at the same time. Machine B fills bottles at a rate of x, which is 1.5 time

s the rate of Machine A. The factory is considering buying a new machine that would replace the other two machines. It can fill at a rate of 3x.
If the factory buys the new machine to replace the other two, which of the following expressions show the increase in rate?
Mathematics
1 answer:
Lana71 [14]4 years ago
3 0

Answer:

\dfrac{4}{3}x

Step-by-step explanation:

Given that:

Rate of bottle filling for Machine B = x

Rate of filling of Machine B is 1.5 times the rate of Machine A.

Let the rate of bottle filling for Machine A = y

As per question statement:

x=1.5y\\\Rightarrow y=\dfrac{2}{3}x

Now, combined rate of both the machines, i.e. A and B = x+y

\Rightarrow x+\dfrac{2}{3}x\\\Rightarrow \dfrac{5}{3}x

Bottle filling Rate of new Machine = 3x

It is given that both the machines are replaced by the new machine.

Now, increase in the rate can be calculated by subtracting the combined rate of both the machines from the bottle filling rate of new machine.

i.e.

3x-\dfrac{5x}3\\\Rightarrow \dfrac{9x-5x}{3}\\\Rightarrow \bold{\dfrac{4x}{3}}

So, the following expression shows the increase in rate:

\dfrac{4}{3}x

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Answer:

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Step-by-step explanation:

hay 60 minutos en una hora y 150 letras por minuto, así que 150 x 60 = 9000

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6 0
3 years ago
solve the problem related to population growth.A city had a population of 23,900 in 2007 and a population of 25,100 in 2012.(a)
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Solution

a)

To find the exponential growth function, we apply the exponential growth formula which is

N(t)=a(1+k)^t

Where

\begin{gathered} a\text{ is the }initial\text{ population} \\ k\text{ is the growth rate} \\ t\text{ is the number of time intervals} \end{gathered}

Given that

\begin{gathered} a=23,900 \\ N(5)=25,100 \\ t=2012-2007=5 \end{gathered}

Substitute the variables into the exponential growth formula

\begin{gathered} 25100=23900(1+k)^5 \\ \text{Divide both sides by 23,900} \\ \frac{25100}{23900}=\frac{23900(1+k)^5}{23900} \\ 1.05021=(1+k)^5^{} \\ \sqrt[5]{1.05021}=1+k \\ 1.00984=1+k \\ \text{Collect like terms} \\ k=1.00984-1 \\ k=0.00984 \end{gathered}

Hence, the exponential growth function is

\begin{gathered} N(t)=23900(1+0.00984)^t_{} \\ N(t)=23900(1.00984)^t \end{gathered}

Hence, the exponential growth function is

N(t)=23900(1.00984)^t

b)

For the population of the city in 2022,

\begin{gathered} t=2022-2007=15 \\ t=15 \end{gathered}

Substitute for t into the exponential growth function

\begin{gathered} N(t)=23900(1.00984)^t \\ N(15)=23900(1.00984)^{15} \\ N(15)=27700\text{ (nearest hundred)} \end{gathered}

Hence, the population is 2022 is 27700 (nearest hundred)

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Answer:

Step-by-step explanation:

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