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3 years ago

Jorge’s printer can print 36 pages in 18 minutes. How many pages can it print in 1 hour

Mathematics

1 answer:

Sloan [31]3 years ago

4 0

Answer:

120 pages can be printed in 1 hour.

Step-by-step explanation:

First divide 36 by 18 to figure out how many pages can be printed in a minute.

Like this: 36 pages ÷ 18 minutes = 2 pages in one minute.

Now, multiply 2 by 60 to figure out how many pages can be printed out in a hour.

Like this: 2 pages in one minute × 60 minutes ( one hour) = 120 pages in one hour can be printed.

Hope this helps!

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Associative property of (21)(7)(3x) ? <br> PLEASE HELPPPP <br> 40 PTS!!!!

sdas [7]

Ehh, now it's letting me answer.

So the associative property is basically as long as it's addition or multiplication, you can move around the groupings and it will still be the same.

For example, you can say:(21 * 7) (3x) instead of (21)(7)(3x)

And it would be equal to the old version.

(To answer your other question, the commutative property is you can move around all the parts of the equation as long as it's addition or multiplication.)

So basically, it would be the first one you sent me.

3 0

3 years ago

Read 2 more answers

For the following amount at the given interest rate compounded continuously, find (a) the future value after 9 years, (b) the

Stella [2.4K]

Answer:

(a) The future value after 9 years is $7142.49.

(b) The effective rate is $r_E=4.759 \:{\%}$ .

(c) The time to reach $13,000 is 21.88 years.

Step-by-step explanation:

The definition of Continuous Compounding is

If a deposit of $P$ dollars is invested at a rate of interest $r$ compounded continuously for $t$ years, the compound amount is

$A=Pe^{rt}$

(a) From the information given

$P=4700$

$r=4.65\%=\frac{4.65}{100} =0.0465$

$t=9 \:years$

Applying the above formula we get that

$A=4700e^{0.0465\cdot 9}\\A=7142.49$

The future value after 9 years is $7142.49.

(b) The effective rate is given by

$r_E=e^r-1$

Therefore,

$r_E=e^{0.0465}-1=0.04759\\r_E=4.759 \:{\%}$

(c) To find the time to reach $13,000, we must solve the equation

$13000=4700e^{0.0465\cdot t}$

$4700e^{0.0465t}=13000\\\\\frac{4700e^{0.0465t}}{4700}=\frac{13000}{4700}\\\\e^{0.0465t}=\frac{130}{47}\\\\\ln \left(e^{0.0465t}\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t\ln \left(e\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t=\ln \left(\frac{130}{47}\right)\\\\t=\frac{\ln \left(\frac{130}{47}\right)}{0.0465}\approx21.88$

3 0

3 years ago

What is 7,5444+789=?

lisov135 [29]

Answer:

76233

Step-by-step explanation:

Add the numbers together:

75444 + 789 = 76233

3 0

3 years ago

Read 2 more answers

PLease help!!!!!!!!!!!!!!!!!!!!!!1

schepotkina [342]

4(5x-20)= -20

20x-80= -20
+80 +80
____________

20x=60
_______
20

x=3

3 0

3 years ago

Read 2 more answers

Situation:A researcher in North America discoversa fossile that contains 65% of its originalamount of C-14..-ktN=NoeNo inital am

zheka24 [161]

SOLUTION

We have been given the equation of the decay as

$\begin{gathered} N=N_0e^{-kt} \\ where\text{ } \\ N_0=initial\text{ amount of C-14 at time t} \\ N=amount\text{ of C-14 at time t = 65\% of N}_0=0.65N_0 \\ k=0.0001 \\ t=time\text{ in years = ?} \end{gathered}$

So we are looking for the time

Plugging the values into the equation, we have

$\begin{gathered} N=N_0e^{-kt} \\ 0.65N_0=N_0e^{-0.0001t} \\ e^{-0.0001t}=\frac{0.65N_0}{N_0} \\ e^{-0.0001t}=0.65 \end{gathered}$

Taking Ln of both sides, we have

$\begin{gathered} ln(e^{-0.000t})=ln(0.65) \\ -0.0001t=ln(0.65) \\ t=\frac{ln(0.65)}{-0.0001} \\ t=4307.82916 \end{gathered}$

Hence the answer is 4308 to the nearest year

8 0

1 year ago