Answer:
See below.
Step-by-step explanation:
I will assume that 3n is the last term.
First let n = k, then:
Sum ( k terms) = 7k^2 + 3k
Now, the sum of k+1 terms = 7k^2 + 3k + (k+1) th term
= 7k^2 + 3k + 14(k + 1) - 4
= 7k^2 + 17k + 10
Now 7(k + 1)^2 = 7k^2 +14 k + 7 so
7k^2 + 17k + 10
= 7(k + 1)^2 + 3k + 3
= 7(k + 1)^2 + 3(k + 1)
Which is the formula for the Sum of k terms with the k replaced by k + 1.
Therefore we can say if the sum formula is true for k terms then it is also true for (k + 1) terms.
But the formula is true for 1 term because 7(1)^2 + 3(1) = 10 .
So it must also be true for all subsequent( 2,3 etc) terms.
This completes the proof.
The histogram is especially useful in comparing mean and median values of a variable. We have that 5.5+6+7+10+7.5+8+9.5+9+8.5+8+7+7.5+6+6.5+5.5=111.5 Since there are 15 values, their mean is 111.5/15=7.43 which is very close to the mean. We also have that 7 onservations are lower than 7.4 while 8 are bigger than 7.4; hence, the diagram is rather balanced and not left-skewed. We cannot tell immediately which one is larger since the values are too close. Any such random process can usually be approximated to a greater or smaller degree by a normal curve; the more points, the better. The histogram shows this (it is kind of a discrete normal curve); all points except 4 will be in this interval of bars.
(1,0) or (2,5) or (3,10)
5-0=5, 10-5=5, and 15-10=5
2x + 12 = 58 <== ur equation (cost of the trip)
2x = 58 - 12
2x = 46
x = 46/2
x = 23 <== price of 1 pass
I believe it would be all of the above let me know if it is right
