7,9,10,11,13 Assume the data list from smallest to largest is A,B,C,D,E. Then from the info given C=10 since 10 is the median. Also the range of the data is 6, so that means that E-A = 6. Since the mean of the data is 10 then (A+B+C+D+E)/5 = 10 or A+B+C+D+E = 50 You can substitute the value for C and E = 6+A (since the range is 6) to get A+B+10+D+(6+A) = 50 or simplified: 2A + B + D =34 Now since the median of the data is 10 the value of B will be less than 10 and A will be less than B. The value of D will be greater than 10 and E will be greater than D. To figure out the data list use trial and error. Start by picking a value of D > 10. Start with 11. From the equation above you have 2A + B + 11 = 34, now pick a value of B < 10. Start with 9 and substitute to solve for A. 2A + 9 + 11 = 34 or 2A = 14 and A=7 So A=7, B=9, C=10, D=11 and E=6+A or 13. To check see if these numbers equal the mean of 50. (7+9+10+11+13)/5 and it does. If it didn't work then you'd have to go back and pick values of B, D again until it does. Hope that helps.
10-d=-34-5d (add 34 to each side) 44-d=-5d (add d to each side) 44=-4d (divide by -4) -11=d
That would mean d times 2.
If you mean 10% then it is 500
If you mean d*r = r*d (where * means multiply_ then it is the commutative property.
10 events in a decathlon
Uh, ten pennies in a dime?
this means 10 pennies in a dime
D = 500, XXX = 10+10+10. Therefore DXXX = 530
Uh, ten pennies in a dime?
No. Decahedron is a 3-D shape with ten faces.
It means not regular or in fractions it means a bigger fraction than a whole (like this 13/10) :D :D
The solution to this ditloid is: 10 Yards for a First Down.
10 Downing Street
M=1000 D=500 CCC= 300 X=10 but the other D can't be placed after X. You may mean V instead of another D at the end. If in this case, MDCCCXV= 1815.
10 dimes in a Dollar
Assuming by "numbers" you mean "whole numbers":{10, 10, 12, 13, 15}If the five numbers are {a, b, c, d, e} with a ≤ b ≤ c ≤ d ≤ e, then:median = 12 → c = 12leaving only two numbers (a, b) ≤ 12.mode = 10 → two or more numbers equal 10 (which is less than 12) → a = b = 10The final two numbers (d, e) are not equal, both greater than 12, and such that the sum of all five numbers is 60.10 + 10 + 12 + d + e = 60 → d + e = 28 → d = 13, e = 15→ {a, b, c, d, e} = {10, 10, 12, 13, 15}[If "numbers" includes "decimal numbers" (ie numbers with a fractional part), then as long as d + e = 28 and 12 < d < e there are infinitely many solutions, eg {10, 10, 12, 12.5, 15.5}, {10, 10, 12, 13.75, 14.25}]