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Amiraneli [1.4K]
2 years ago
6

Question 10. Suppose U and W are subspaces of V, dim V = n, dim U + dim W = n, and U NW = {0}. Prove that V=UW.

Mathematics
1 answer:
Bess [88]2 years ago
4 0

Answer: If the dimension of V is n, then V has n elements.

Now, dim(U) + dim(W) = n, this means that the addition of the dimensions of U and W also has n elements.

and because U and W are subspaces of V, you know that every element on U and W is also an element of V.

If U ∩ W = ∅, means that there are no elements in common between U and W.

Because there are no elements in common, then Dim(U) + Dim(W) = dim(U ∪ W) = n

So U ∪ W has the same number of elements as V, and every element of W and U is also an element of V

this means that U ∪ W = V.

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PLEASE ANSWER CORRECTLY
inn [45]

Answer:

$76.32

Step-by-step explanation:

First, add up all the costs for everything:

17 + 30 + 25 = 72

Now, we need to add the sales tax. To do this we multiply 72 by 1.06. This is because the sales tax is 6% (the decimal form of 6% is 0.06), and we add it on to the 100% cost (which is 1 in decimal form).

(72)(1.06) = 76.32

Therefore the total cost of everything Walter bought plus the tax rate is $76.32.

<em>I hope this helps!!</em>

<em>- Kay :)</em>

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2 years ago
How many copies of 1 twelfth are equivalent to 3 sixths? Help!
andrezito [222]

Answer:

6/12 is equal to 3/6 ( 6 twelfths is equal to 3 sixths )

Step-by-step explanation:  

multiply both sides of 3/6 by 2 to get the same denominator, doing that will get you 6/12

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3 years ago
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Step-by-step explanation:

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Which of the following is the MOST resistant to erosion?
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A person purchased a slot machine and tested it by playing it 1,137 times. There are 10 different categories of outcomes, includ
ivann1987 [24]

Answer:

p_v= P(\chi^2_{9}>11.517)=0.2419

And on this case if we see the significance level given \alpha=0.1 we see that p_v>alpha so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

Step-by-step explanation:

A chi-square goodness of fit test determines if a sample data obtained fit to a specified population.

p_v represent the p value for the test

O= obserbed values

E= expected values

The system of hypothesis for this case are:

Null hypothesis: O_i = E_i[/tex[Alternative hypothesis: [tex]O_i \neq E_i

The statistic to check the hypothesis is given by:

\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

On this case after calculate the statistic they got: \chi^2 = 11.517

And in order to calculate the p value we need to find first the degrees of freedom given by:

df=n-1=10-1=9, where k represent the number of levels (on this cas we have 10 categories)

And in order to calculate the p value we need to calculate the following probability:

p_v= P(\chi^2_{9}>11.517)=0.2419

And on this case if we see the significance level given \alpha=0.1 we see that p_v>alpha so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

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