Danny reads 4 pages of a book each day.After 12 days, hes still has 82 pages left to read.How many pages does the book have?

Answer: Well, he reads 48 pages in those 12 days, that is 12*4 pages.
He has already read 48 pages and he still has 82, so he has a 130=48+82 page book.

Related Questions

You are paid \$78 for 6 and a half hours of work. What is your rate of pay?

The most important part of the problem lies in the fact that the hours have to be converted to minutes before starting to solve the problem.
1 hour = 60 minutes
6 and half hours = (6 * 60) + 30 minutes
= 360 + 30
= 390 minutes
So the payment for 390 minutes is = 78 dollars
Then
The payment for 60 minutes = (78/390) * 60
= 2 * 6 dollars
= 12 dollars
So the rate of pay given is \$12 per hour. I hope the procedure is easy enough for you to understand. You can always use this method for solving similar type of problems.
\$78/ (6 1/2 hours)= \$12/hour.

The rate of pay is \$12 per hour~

9 1/2 gal = ? qt?

-In US 1 gal= 4qts.
-In order to find out how much 9 1/2 gal equals in quarts, you need to multiply 9 1/2 by 4.
-To make it easier you can turn 9 1/2 to 9.5
- So the ANSWER is 38 quarts!!!!
- 9 1/2= 38 qt
-HOPE THAT HELPED!!!!
Hey there! 9 1/2 in decimal form is 9.5 because 1/2=0.5 and then you add 9 to 0.5 (9+0.5=9.5). So, 9 1/2 gallons equals 38 quarts.

Four liters or one gallon which is bigger

4 liters because 4 liters equals 1.05 gallons and a gallon equals 3.78 liters

Set 1: {0, 2, 4, 6, 8, 10}Set 2: {1, 3, 5, 7, 9, 11}
what is the means to mad ratio?
are they similar?

To find the mean absolute deviation of the data, start by finding the mean of the data set. Find the sum of the data values, and divide the sum by the number of data values. Find the absolute value of the difference between each data value and the mean: |data value – mean|.

Step-by-step explanation:

Calculate Mean Absolute Deviation (M.A.D)

A website captures information about each customer's order. The total dollar amounts of the last 8 orders are listed in the table below. What is the mean absolute deviation of the data?

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