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Dima020 [189]
3 years ago
14

The accompanying specific gravity values describe various wood types used in construction. 0.320.350.360.360.370.380.400.400.40

0.410.410.420.420.420.420.420.430.44 0.450.460.460.470.480.480.490.510.54 0.540.550.580.630.660.660.670.680.78 Construct a stem-and-leaf display using repeated stems. (Enter numbers from smallest to largest separated by spaces. Enter NONE for stems with no values.)
Engineering
1 answer:
marysya [2.9K]3 years ago
8 0

Answer:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\  \ \\ {0.7} & {\vert} & {8} \ \ \end{array}

Explanation:

Given

0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,

0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,

0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,

0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.

0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,

0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.

0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.

0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.

0.78.

The 0.3's is will be plotted as thus:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}

The 0.4's is as follows:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}

The 0.5's is as follows:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}

The 0.6's is as thus:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}

Lastly, the 0.7's is as thus:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.7} & {\vert} & {8} \ \ \end{array}

The combined steam and leaf plot is:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\  \ \\ {0.7} & {\vert} & {8} \ \ \end{array}

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Answer:

Explanation:

Step 1:

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     gcd(m,n) = gcd (n,m mod n)

the gcd(31415,14142) by applying Euclid's algorithm is

    gcd(31,415,14,142) =gcd(14,142,3,131)

                                  =gcd=(3,131, 1,618)

                                   =gcd(1,618, 1,513)

                                   =gcd(1,513, 105)

                                   =gcd(105, 43)

                                    =gcd(43, 19)

                                     =gcd(19, 5)

                                      =gcd(5, 4)

                                      =gcd(4, 1)

                                      =gcd(1, 0)

                                      =1

STEP 2

b)  The number of comparison of given input with the algorithm based on  checking consecutive integers and Euclid's algorithm is

     The number of division using Euclid's algorithm =10 from part (a)

      The consecutive integer checking algorithm:

      The number of iterations =14,142 and 1 or 2 division of iteration.

        14,142 ∠= number of division∠ = 2*14,142

         Euclid's algorithm is faster by at least 14,142/10 =1400 times

          At most 2*14,142/10 =2800 times.

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Answer:

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