Answer:

Step-by-step explanation:
To find x₁ and x₂ :
![\left[\begin{array}{ccc}-4&1\\5&4\\\end{array}\right] \times \left[\begin{array}{ccc}x_1\\x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%261%5C%5C5%264%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%5C%5Cx_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%5C%5C-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step 1
Multiply first 2 x 2 matrix with 2 x 1 vector, we get
![\left[\begin{array}{ccc}-4x_1&+ x_2\\5x_1&+ 4x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4x_1%26%2B%20%20x_2%5C%5C5x_1%26%2B%20%204x_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%5C%5C-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5Cend%7Barray%7D%5Cright%5D)
Step 2
Add the 2 x 1 matrices on LHS, we get
![\left[\begin{array}{ccc}-4x_1&+x_2&+11\\5x_1&+4x_2&-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4x_1%26%2Bx_2%26%2B11%5C%5C5x_1%26%2B4x_2%26-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5Cend%7Barray%7D%5Cright%5D)
Step 3,
we get

and

Step 4,
Simplify, we get

Step 5,
multiply eqn(1) by 4
we get

Step 6,
eqn (2) - eqn(3)
we get

substituting in eqn (1), we get

so, we get

Therefore

54 that's going to be your answer to your question
I would say C ejejdkidjejdkdndkeiej
Answer:
The equation y = 3x shows a proportional relationship between x and y
Step-by-step explanation:
- <em>The proportional relationship is represented graphically by a line passes through the </em><em>origin point </em>
- <em>The equation of the proportional relationship is </em><em>y = m x</em>
Let us solve the question
Substitute x by 0 to find the value of y, if y = 0, then the line represents this equation passes through the origin point, then the equation shows a proportional relationship
∵ y = 3x
∵ x = 0
∴ y = 3(0)
∴ y = 0
→ That means the line that represents the equation passes through
the origin point
∴ The equation y = 3x shows a proportional relationship between x and y
Answer:
a) 99.97%
b) 65%
Step-by-step explanation:
• 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.
• 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.
• 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.
Mean of 98.35°F and a standard deviation of 0.64°F.
a. What is the approximate percentage of healthy adults with body temperatures within 3 standard deviations of the mean, or between 96.43°F and 100.27°F?
μ - 3σ
98.35 - 3(0.64)
= 96.43°F
μ + 3σ.
98.35 + 3(0.64)
= 100.27°F
The approximate percentage of healthy adults with body temperatures is 99.97%
b. What is the approximate percentage of healthy adults with body temperatures between 97 .71°F and 98.99°F?
within 1 standard deviation from the mean - that means between μ - σ and μ + σ.
μ - σ
98.35 - (0.64)
= 97.71°F
μ + σ.
98.35 + (0.64)
= 98.99°F
Therefore, the approximate percentage of healthy adults with body temperatures between 97.71°F and 98.99°F is 65%