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slega [8]
2 years ago
15

Solve the system of equations please 1/4x+y=1 3/2x-y=4/3

Mathematics
1 answer:
Inessa [10]2 years ago
4 0

Answer:

\huge\boxed{x=\dfrac{4}{3};\ y=\dfrac{2}{3}\to\ \left(\dfrac{4}{3};\ \dfrac{2}{3}\right)}

Step-by-step explanation:

\underline{+\left\{\begin{array}{ccc}\dfrac{1}{4}x+y=1\\\dfrac{3}{2}x-y=\dfrac{4}{3}\end{array}\right}\qquad|\text{add both sides of the equations}\\\dfrac{1}{4}x+\dfrac{3}{2}x=1+\dfrac{4}{3}\\\\\dfrac{1}{4}x+\dfrac{3\cdot2}{2\cdot2}x=\dfrac{3}{3}+\dfrac{4}{3}\\\\\dfrac{1}{4}x+\dfrac{6}{4}x=\dfrac{3+4}{3}\\\\\dfrac{1+6}{4}x=\dfrac{7}{3}\\\\\dfrac{7}{4}x=\dfrac{7}{3}\qquad|\text{multiply both sides by}\ \dfrac{4}{7}

\dfrac{4\!\!\!\!\diagup}{7\!\!\!\!\diagup}\cdot\dfrac{7\!\!\!\!\diagup}{4\!\!\!\!\diagup}x=\dfrac{4}{7\!\!\!\!\diagup}\cdot\dfrac{7\!\!\!\!\diagup}{3}\\\\x=\dfrac{4}{3}

Put it to the first equation

\dfrac{1}{4\!\!\!\!\diagup}\cdot\dfrac{4\!\!\!\!\diagup}{3}+y=1\\\\\dfrac{1}{3}+y=1\qquad|\text{subtract}\ \dfrac{1}{3}\ \text{from both sides}\\\\y=\dfrac{3}{3}-\dfrac{1}{3}\\\\y=\dfrac{3-1}{3}\\\\y=\dfrac{2}{3}

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The following results come from two independent random samples taken of two populations.
photoshop1234 [79]

Answer:

(a)\ \bar x_1 - \bar x_2 = 2.0

(b)\ CI =(1.0542,2.9458)

(c)\ CI = (0.8730,2.1270)

Step-by-step explanation:

Given

n_1 = 60     n_2 = 35      

\bar x_1 = 13.6    \bar x_2 = 11.6    

\sigma_1 = 2.1     \sigma_2 = 3

Solving (a): Point estimate of difference of mean

This is calculated as: \bar x_1 - \bar x_2

\bar x_1 - \bar x_2 = 13.6 - 11.6

\bar x_1 - \bar x_2 = 2.0

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)

(1.0542) \to (2.9458)

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}

2.0 \± 1.1270

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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Answer:

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