1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vichka [17]
3 years ago
12

Karyn proofreads 15 pages in 2 hours for $40.

Mathematics
2 answers:
OLEGan [10]3 years ago
8 0
Hi there.

Since she reads 15 pages in 2 hours, all we have to do to find out how much she reads in 1 hour is divide.

15/2 = 7.5

7.5 pages/hour

Same applies to the pay

40/2 = 20

$20/hour

Karyn proofreading rate is 7.5 pages per hour. Karyn receives on average $20 per hour.

~
svetlana [45]3 years ago
7 0
15/2 is her rate per hour... she proofreads 7 and a half pages in 1 hour and for 20 dollars 

20 / 7.5  is how much does she make per page or 40/15 
she makes roughly $ 2 .66
You might be interested in
What can I add to get -6?
krok68 [10]
So what you would need to add to get -6 is -5-1 then your answer would be -6 hope this helps!
3 0
2 years ago
Read 2 more answers
Find the area of the following composite figure please ​
telo118 [61]

Step-by-step explanation:

answer is in photo above

5 0
2 years ago
A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon
fiasKO [112]

Answer:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.

This means that n = 603, \pi = \frac{142}{603} = 0.2355

95% confidence level

So \alpha = 0.05, z is the value of Z that has a p-value of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

8 0
2 years ago
What is the Oder of 7/10 7/8 7/12 7/9
Flauer [41]
Least to Greatest Order -

   7/12,   7/10,   7/9,   7/8
(Least)                    (Greatest)
5 0
3 years ago
Read 2 more answers
Please help me with this question I also need to see the steps if possible!
umka21 [38]

Hi there!

To solve, we must use the following trig identity:

sin(u - v) = sin(u)cos(v) - sin(v)cos(u)

We can rewrite the left hand side of the equation as:

\frac{sin(u)cos(v)-sin(v)cos(u)}{sin(u)cos(v)}

Split the fraction:

\frac{sin(u)cos(v)}{sin(u)cos(v)} - \frac{sin(v)cos(u)}{sin(u)cos(v)} =

First fraction reduces to 1:

1 - \frac{sin(v)cos(u)}{sin(u)cos(v)} =

Simpify each with common arguments:

1 - tan(v)cot(u)

4 0
3 years ago
Other questions:
  • 5. What number is 45% of 200?<br>​
    9·2 answers
  • -3 and 12 find the midpoint?​
    15·1 answer
  • Square root of 98k???
    8·1 answer
  • A(n) _____is a integer less than zero.​
    15·2 answers
  • How do you write this equation x+5=2(7+x)
    8·2 answers
  • Narra situasiones de la vida rreal las cuales tienen la utilidad los numer9s enteros​
    7·1 answer
  • Find the value of x in the triangle shown below.
    10·1 answer
  • 460 000 X 500 000<br> In Scientific notation
    10·1 answer
  • HELP ME PLEASEEEE<br><br> thank you &lt;3
    12·1 answer
  • Quick algebra 1 question for 50 points!
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!