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oksano4ka [1.4K]
3 years ago
12

Determine all real values of p such that the set of all linear combination of u = (3, p) and v = (1, 2) is all of R2. Justify yo

ur answer.
Mathematics
1 answer:
Rama09 [41]3 years ago
7 0

Answer:

p ∈ IR - {6}

Step-by-step explanation:

The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2

is all R2 ⇔

u\neq 0_{R2}      

v\neq 0_{R2}

And also u and v must be linearly independent.

In order to achieve the final condition, we can make a matrix that belongs to R^{2x2} using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.

Let's make the matrix :

A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]

We used the first vector ''u'' as the first column of the matrix A

We used the  second vector ''v'' as the second column of the matrix A

The determinant of the matrix ''A'' is

Det(A)=6-p

We need this determinant to be different to zero

6-p\neq 0

p\neq 6

The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that p\neq 6

We can write : p ∈ IR - {6}

Notice that is p=6 ⇒

u=(3,6)

v=(1,2)

If we write 3v=3(1,2)=(3,6)=u , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.

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Answer:

1) Null hypothesis:\mu \leq 500  

Alternative hypothesis:\mu > 500  

z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90  

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

z_{crit}= 2.33

For this case we see that the calculated value is higher than the critical value

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

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

p_v =P(z>5.90)=1.82x10^{-9}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

Step-by-step explanation:

Part 1

Data given

\bar X=640 represent the sample mean

\sigma=150 represent the population standard deviation

n=40 sample size  

\mu_o =500 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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

Step1:State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 500, the system of hypothesis would be:  

Null hypothesis:\mu \leq 500  

Alternative hypothesis:\mu > 500  

Step 2: Calculate the statistic

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

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

z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90  

Step 3: Calculate the critical value

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

z_{crit}= 2.33

Step 4: Compare the statistic with the critical value

For this case we see that the calculated value is higher than the critical value

Step 5: Decision

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

Part 2

P-value  

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

p_v =P(z>5.90)=1.82x10^{-9}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

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