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vaieri [72.5K]
4 years ago
10

Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial t

ransport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known. The data obtained are as follows:
n_1 = 10
x_1 = 87.6
σ_1 = 1
n_2 = 12
x^2 = 74.5
σ_2 = 1.5.

Required:
If μ _1 and μ _2 denote the true mean tensile strengths for the two grades of spars. Construct a 90 percentage confidence interval on the difference in mean strength.
Mathematics
1 answer:
SCORPION-xisa [38]4 years ago
5 0

Answer:

(12.141, 14.059)

Step-by-step explanation:

Explanation is provided in the attached document.

Download docx
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Step-by-step explanation:

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An electronics company produces​ transistors, resistors, and computer chips. Each transistor requires 3 units of​ copper, 2 unit
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Answer:

475 transistors, 25 resistors and 50 computer chips can be produced.

Step-by-step explanation:

Let us consider, p = Number of transistors.

                           q = Number of resistors.

                            r = Number of computer chips.

The following three linear equations according to question,

3\times p + 3\times q + 2\times r = 1600\\2\times p + 1\times q + 1\times r = 1025\\1\times p + 2\times q + 2\times r = 625

The matrix form of any system, Ax = B

Where, A = Coefficient matrix

            B = Constant vector

            x = Variable vector

A = \left[\begin{array}{ccc}3&3&2\\2&1&1\\1&2&2\end{array}\right], x = \left[\begin{array}{ccc}p\\q\\r\end{array}\right], B = \left[\begin{array}{ccc}1600\\1025\\625\end{array}\right]

The inverse matrix, A^{-1} can be found by using the following formula,

           A^{-1} = \frac{1}{det A}\times (C_{A}) ^{T}

Where, det A = Determinant of matrix A.

                 C_{A} = Matrix of cofactors of A

Now, applying this formula to find A^{-1};

det A = \left[\begin{array}{ccc}3&3&2\\2&1&1\\1&2&2\end{array}\right] = 3\times(2-2)-3\times(4-1)+2\times(4-1) = -3

Here, det A\neq 0, thus the matrix is invertible.

C_{A} = \left[\begin{array}{ccc}(2-2)&-(4-1)&(4-1)\\-(6-4)&(6-2)&-(6-3)\\(3-2)&-(3-4)&(3-6)\end{array}\right] = \left[\begin{array}{ccc}0&-3&3\\-2&4&-3\\1&1&-3\end{array}\right] \\(C_{A}) ^{T} = \left[\begin{array}{ccc}0&-3&3\\-2&4&-3\\1&1&-3\end{array}\right] ^{T} = \left[\begin{array}{ccc}0&-2&1\\-3&4&1\\3&-3&-3\end{array}\right]

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So, p = 475, q = 25, r = 50.      

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