The answer at this website is incorrect....
Yay! Thanks!
109/2=54.5, so that means the 54th and the 55th will be the median.
Therefore, the greatest possible number of students is 30.
Okay. We need to find out the biggest number that can go in the middle of the Venn diagram.
As I get, the solution is 2.
Is this correct?
20 40 60 80 100 120 140 160 180 200 240 260 280 300 320 340 360 380 400 420 460 480 500 520 540 560 580 600 620 640 680 700 720 740 760 780 800 820 840 860 900 920 940 960 980 Total = 45
A little hint.....
The numbers that will work are:
100, 120, 140, 160, 180, 200, 240...
Keep counting by 20, but see what numbers don't work!
The ways to paint the red faces are:
1 and 2
1 and 3
1 and 4
1 and 5
2 and 3
2 and 4
3 and 5
3 and 6
2 and 6
4 and 6
4 and 5
5 and 6
So 12 ways. I see my mistake!
Really? I'm sorry... I will try again.
This has the solution to this problem:
https://slideplayer.com/slide/10345148/
Solution:
Let's list out the ways that the red faces don't add up to 7.
1+3
1+2
1+4
1+5
2+3
2+4
So, that is 6 ways that this works.
Please tell me if I am correct.
If altitude CD = √3 in centimeters, what is the number of square centimeters in the area of ΔABC?
a. 10yrs=1 decade so, 1.34 to the 1/10 gives you 1.03 {1.34^1/10}=1.03 YEARLY GROWTH FACTOR
3.93 x1.03^y FORMULA
b. 1.34^10=18.67 CENTURY GROWTH FACTOR=18.67
so 3.93 x 18.67^c FORMULA You raise the 1.34 by 10 because there are 10 decades in 1 century.
I think it says "a." Can you make sure?
This question was also posted here, with a very reasonable explanation. https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.770190.html
Eastward with head start) = 8 hrs * x m/hr
Westward = 4 hrs * (x+10) m/hr summed = 400
8x + 4x+40 = 400
x = 30 m/hr the westbound is traveling at 40 m/hr
Some parts of the question are missing. If you fix the question, I would love to help :D
