Answer:
150 minutes
Step-by-step explanation:
If Anna read 75% of her 300-page book, then she has already read:
300 × 75%
= 300×0.75
= 225
225 pages.
If Anna has already read 225 pages, then she has:
300 - 225 = 75
75 pages left still to read.
If Anna reads 2 minutes per page, then she has
75 × 2 = 150 minutes.
She has 150 minutes of reading left to finish her book.
By the way, it's weird to say you have 150 minutes of something to do. In real life 150 minutes = 2 1/2 hours. jfyi. Have a great day!
Answer: x ≤ 15
Step-by-step explanation: All we need to do is divide both sides by 1/3, so1/3x / 1/3 and 5/ 1/3 = 15. So x ≤ 15.
Hope this helps!
7 1/2 years ago to be precise
If correct brainley please
Thank you
Answer: The required system of equations representing the given situation is

Step-by-step explanation: Given that Sam needs to make a long-distance call from a pay phone.
We are to write a system to represent the situation.
Let x represent the number of minutes Sam talked on the phone and y represents the total amount that he paid for the call.
According to the given information,
with prepaid phone card, Sam will be charged $1.00 to connect and $0.50 per minute.
So, the equation representing this situation is

Also, if Sam places a collect call with the operator he will be charged $3.00 to connect and $0.25 per minute.
So, the equation representing this situation is

Thus, the required system of equations representing the given situation is

Answer:
7/15 km
Step-by-step explanation:
Find the least common denominator between the two fractions. The least common denominator in this case would be 15, so you have to multiply 1/3 by 5 to have a denominator of 15. Then you multiply 1/5 by 3 to have a denominator of 15. You will find that Mark ran a distance of 2 5/15 km and Shaun ran 3 3/15 km. To make it even easier, you can add the whole number back into the fraction so Mark will have ran 35/15 km and Shaun will have ran 48/15 km meaning that Shaun will have ran more. Then you set you fractions up and subtract them.