Ask question

Ask question

All categories

2 years ago

7

A clerk at Agency Y has determined that at speeds of 25 pages per minute it is typical for 3 out of every 100 pages she photocop

ies to contain some type of printing error. How many pages with printing errors would she expect to find during a 4-hour photocopying session at 25 pages per minute

1 answer:

nalin [4]2 years ago

3 0

Answer:

180 pages

Step-by-step explanation:

Since her photocopy session is 25 pages per minute, the total number of pages photocopied in 4 hour session can be calculated as;

number of pages = time × speed

= (4 × 60) × 25

= 240 ×25

= 6000 pages

Thus, number of pages photocopied in 4 hours is 6000.

For every 100 pages, 3 would contain some type of printing error. The number of pages she expect with printing error during 4 hour session can be determined as;

= $\frac{6000}{100}$ × 3

= 60 × 3

= 180 pages

The expected number of pages with printing error during 4 hour photocopying session is 180 pages.

You might be interested in

NEED HELP ASAP!!! for geometry class

balandron [24]

Answer:

B) AAS

Step-by-step explanation:

B) AAS

5 0

2 years ago

Read 2 more answers

PLEASE HELP ILL GIVE A BRAINLIEST

zavuch27 [327]

Guessing you meant by a factor of 3... not 13 since none of the solutions are anywhere close to that.

to compress is to smoosh.
It's the equation of a line. If you smoosh it vertically the line becomes flatter and the slope smaller.

Essentially g(x) = 3f(x)
3y = 12x + 6
y = (12/3)x + 6/3
y = 4x + 2

8 0

2 years ago

Paulo works at the United Nations. He researched what percent of the world's population lives on each continent. He surveys a sa

Fofino [41]

Answer:

confidence interval using a two sample t test between percents

Step-by-step explanation:

confidence interval using a two sample t test between percents This can be used to compare percentages drawn from two independent samples in this case employees. It is used to compare two sub groups from a single sample example the population on the planet

5 0

2 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

2 years ago

Point C is located in which quadrant?<br><br> a.) I<br> b.) II<br> c.) III<br> d.) IV

Aleonysh [2.5K]

B, it is located on the 2nd quadrant

6 0

2 years ago

Read 2 more answers