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Artyom0805 [142]
2 years ago
6

Ax +by = 16

Mathematics
1 answer:
Talja [164]2 years ago
3 0

Answer:

  • The system has infinitely many solutions when a = 3/4 and b = 1/2

<h3>Given </h3>

System of linear equations

  • ax +by = 16
  • 3x+2y=64

<h3>To find </h3>

  • Values of a and b that leads to infinitely many solutions

<h3>Solution</h3>

The linear system of equations has infinitely many solutions when lines overlap, so both have same slope and y-intercept.

We can solve it in two different ways

1. Note that when the first equation is multiplied by 4, it has same value of constant as the second equation, which is 64.

  • 4ax + 4by = 64
  • 3x + 2y = 64

When compared it gives us:

  • 4a = 3 ⇒ a = 3/4
  • 4b = 2 ⇒ b = 2/4 = 1/2

2. Convert both equations from standard to slope-intercept form and compare:

  • ax + by = 16
  • by = -ax + 16
  • y = - (a/b)x + 16/b

  • 3x + 2y = 64
  • 2y = - 3x + 64
  • y = - (3/2)x + 32

Work out values of a and b:

  • 16/b = 32 ⇒ b = 16/32 ⇒ b = 1/2
  • - (a/b) = - 3/2 ⇒ a/b = 3/2 ⇒ a = 3b/2 ⇒ a = 3(1/2)/2 ⇒ a = 3/4
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Answer:

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(c)\ CI = (0.8730,2.1270)

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Given

n_1 = 60     n_2 = 35      

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\bar x_1 - \bar x_2 = 13.6 - 11.6

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Solving (b): 90% confidence interval

We have:

c = 90\%

c = 0.90

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.90

\alpha = 0.10

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.10/2}

z_{\alpha/2} = z_{0.05}

The z score is:

z_{\alpha/2} = z_{0.05} =1.645

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.645 * \sqrt{0.0735+0.2571}

2.0 \± 1.645 * \sqrt{0.3306}

2.0 \± 0.9458

Split

(2.0 - 0.9458) \to (2.0 + 0.9458)

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Hence, the 90% confidence interval is:

CI =(1.0542,2.9458)

Solving (c): 95% confidence interval

We have:

c = 95\%

c = 0.95

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.95

\alpha = 0.05

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.05/2}

z_{\alpha/2} = z_{0.025}

The z score is:

z_{\alpha/2} = z_{0.025} =1.96

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.96 * \sqrt{0.0735+0.2571}

2.0 \± 1.96* \sqrt{0.3306}

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Split

(2.0 - 1.1270) \to (2.0 + 1.1270)

(0.8730) \to (2.1270)

Hence, the 95% confidence interval is:

CI = (0.8730,2.1270)

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