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ehidna [41]
2 years ago
14

Solve the system: -5x+6y=55 4x+3y=34 Write down your solution.

Mathematics
1 answer:
svetlana [45]2 years ago
4 0

Answer:

(1, 10)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

<u>Algebra I</u>

  • Solving systems of equations using substitution/elimination

Step-by-step explanation:

<u>Step 1: Define Systems</u>

-5x + 6y = 55

4x + 3y = 34

<u>Step 2: Rewrite Systems</u>

4x + 3y = 34

  1. Multiply everything by -2:                    -8x - 6y = -68

<u>Step 3: Redefine Systems</u>

-5x + 6y = 55

-8x - 6y = -68

<u>Step 4: Solve for </u><em><u>x</u></em>

<em>Elimination</em>

  1. Combine equations:                    -13x = -13
  2. Divide -13 on both sides:             x = 1

<u>Step 5: Solve for </u><em><u>y</u></em>

  1. Define equation:                    4x + 3y = 34
  2. Substitute in <em>x</em>:                       4(1) + 3y = 34
  3. Multiply:                                  4 + 3y = 34
  4. Isolate <em>y</em> term:                        3y = 30
  5. Isolate <em>y</em>:                                 y = 10
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Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

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Since is a two-sided test the p value would be:  

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Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

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Answer:

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Before Ray gave Jan cards, she had 18,

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