Answer: Can you upload a picture of the question?
Step-by-step explanation:
Answer:
Step-by-step explanation:
Assuming Roberto wants to completely fill each page that he puts cards in, this function describes the number of 2-card pages, a, and 3-card pages, b.
2a + 3b =18
Ricardo can fill up 9 2-card pages, and 6 3-card pages.
a=9, b=0
We must add 2 3-card pages at a time,so that we have an even number for the 2-card pages:
a=6, b=2
Add 2 to b once more:
a=3, b=4
One more time:
a=0, b=6:
Thus, Ricardo can display his figures in the following page combinations:
a=9, b=0
a=6, b=2
a=3, b=4
a=0, b=6
Remember that a= number of 2-card pages and b=number of 3-card pages
There are 4 different ways that Ricardo can arrange his figures in terms of what kind of pages he uses.
Answer:
d = 10√10 ≈ 31.62
Step-by-step explanation:
d = √l²+w²+h²
d = √5²+8²+2²
d = √25+81+4
d = √110
d = 10*√10
Answer:
7,200 words
Step-by-step explanation:
The chance of student 1's birthday being individual is 365/365 or 100%.
Then the chance of student 2's birthday being different is 364/365.
Then it's narrowed down to 363/365 for student 3 and so on until you get all 10 students.
If you multiply all these values together, the probability would come out at around 0.88305182223 or 0.88.
To get all the same birthday you'd have to the chance of one birthday, 1/365 and multiply this by itself 10 times. This will produce a very tiny number. In standard form this would be 2.3827x10'-26 or in normal terms: 0.23827109210000000000000000, so very small.
