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2 years ago

What is the solution to this system of equations?

Mathematics

1 answer:

damaskus [11]2 years ago

3 0

Since this is multiple choice ...

• check if B is correct:

x = 0, y = 4, z = 1 ⇒ 3x - y + 4z = -4 + 4 = 0 ≠ -10

(it's not)

• check if C is correct:

x = -2, y = 4, z = 0

⇒ 3x - y + 4z = -6 - 4 + 0 = -10

⇒ 3x - y + 4z = 2 + 4 + 0 = 6

⇒ 3x - y + 4z = -4 - 4 + 0 = -8

While this solution does satisfy the system, it can still have infinitely many other solutions that would work.

• check if D is correct:

Eliminate one of the variables from each equation. For instance,

(3x - y + 4z) + (-x + y + 2z) = -10 + 6

2x + 6z = -4

x + 3z = -2

and

(2x - y + z) + (-x + y + 2z) = -8 + 6

x + 3z = -2

but now it's impossible to eliminate one of the variables.

Therefore there are infinitely many solutions to the system [D].

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Step-by-step explanation:

1) Data given and notation

$\bar X$ represent the sample mean

$s$ represent the standard deviation for the sample

$n$ sample size

$\mu_o =20$ 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)

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Null hypothesis: $\mu \leq 20$

Alternative hypothesis: $\mu > 20$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

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$df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

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First we need to calculate the degrees of freedom given by:

$df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

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$df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

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$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

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