Ex-2.19

(define (cc amount coin-values)
  (cond ((= amount 0) 1)
        ((or (< amount 0) (no-more? coin-values)) 0)
        (else
         (+ (cc amount
                (except-first-denomination coin-values))
            (cc (- amount
                   (first-denomination coin-values))
                coin-values)))))
(define (first-denomination coins)
  (car coins))
(define (no-more? coin-values)
  (= (length coin-values) 0))
(define (except-first-denomination coin-values)
  (cdr coin-values)) 

Order of coin-values doesn't affect the answer produced, because in the cc we're treating coin-values as just a set of values without any assumption of its order.

Ex-2.18

(define (reverse l)
  (define (iter a r)
    (if (null? a) r
        (iter (cdr a) (cons (car a) r))))
  (iter l '()))

Ex-2.17

(define (last-pair l)
  (list (list-ref l (- (length l) 1))))

Ex-2.16

passed

Ex-2.15

Yes, She is right. And the reason is, more the uncertain number is repeated more uncertainity it brings to the system(overall calculation).

Ex-2.14

(define (par1 r1 r2)
  (div-interval (mul-interval r1 r2)
                (add-interval r1 r2)))
(define (par2 r1 r2)
  (let ((one (make-interval 1 1)))
    (div-interval one
                  (add-interval (div-interval one r1)
                                (div-interval one r2))))) 

Indeed the results are different:
1 ]=> (par1 (make-interval 3 4) (make-interval 5 7))
;Value 1: (1.3636363636363638 . 3.5)
1 ]=> (par2 (make-interval 3 4) (make-interval 5 7))
;Value 2: (1.875 . 2.5454545454545454)

with center-percent form also the results are different
1 ]=> (par1 (make-center-percent 5 10) (make-center-percent 10 5))
;Value 4: (2.671875 . 4.125)
1 ]=> (par2 (make-center-percent 5 10) (make-center-percent 10 5))
;Value 5: (3.053571428571429 . 3.609375)

Ex-2.13-todo

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