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3 years ago

Please help me its about solving system by substitution

Mathematics

1 answer:

Andrew [12]3 years ago

6 0

Answer:

(-3, 4)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtract Property of Equality

Algebra I

Terms/Coefficients
Solving systems of equations using substitution/elimination

Step-by-step explanation:

Step 1: Define Systems

y = -x + 1

2x + 3y = 6

Step 2: Solve for x

Substitution

Substitute in y: 2x + 3(-x + 1) = 6
Distribute 3: 2x - 3x + 3 = 6
Combine like terms: -x + 3 = 6
Isolate x terms: -x = 3
Isolate x: x = -3

Step 3: Solve for y

Define equation: y = -x + 1
Substitute in x: y = -(-3) + 1
Simplify: y = 3 + 1
Add: y = 4

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$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

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Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

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

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

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