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Colt1911 [192]
3 years ago
14

A checkbook balance sheet shows an initial balance for the month of $300. During the month, checks were written in the amounts o

f $25, $82, $213, and $97. Deposits were made into the account in the amounts of $84 and $116. What was the balance at the end of the month?
Mathematics
1 answer:
DaniilM [7]3 years ago
5 0

Answer:

your answer  is 200 if you add 84 and 116 but is you add 25 82 213 and 97 it will be 417

Step-by-step explanation:

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4 multiplied by 3.99 will = your answer....Please mark brainliest...
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Determine if the following statement is true or false. (Enter TRUE if the statement is true, enter a counterexample if the state
ehidna [41]

Answer:

The answer is: the satement is FALSE.  

Step-by-step explanation:

The first step is to transform the propositons in the statement into symbols:

A: m is any positive integer

B: n is any positive integer

C: mn is a a perfect square

D: m is a perfect square

E: n is a perfect square

After that we replace the symbols into the statement:

If A and B and C, then D and E

Logical connectives on the satement are:

  • p⇒q conditional (if ...,then...). For this connective the only case when is FALSE is when the antecedent is true and the consequent is false.  
  • p ∧ q conjunction (p and q). For this connective the only case when is TRUE is when the two propositions are true at the same time.  

So now, we are going to replace the logical connectives on the statement:

((A ∧ B) ∧ C)  ⇒ (D ∧ E)

There is a hierarchy in logical connectives, first we evaluate those inside brackets. The principal connective is evaluated at the end. In this particular case, the principal connective is the conditional.

As we have five different propositions, the combinations are given by the following rule:

2^n= 2^5=32

32 different combinations  

In figure 1, added bellow, it shows the construction of truth table.

In figure 2, added bellow, it shows the development of the truth table. In the development, we don't have as a result a tautology, therefore the statement is FALSE.  

Download pdf
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>
8 0
3 years ago
G with the definition of covariance, prove cov[(ay − b),(cy − d)] = accov(x, y ), where x, y are random variables and a, b, c, d
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By definition of covariance,


\mathrm{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]


\mathrm{Cov}(X,Y)=\mathbb E[XY-\mathbb E[X]Y-X\mathbb E[Y]+\mathbb E[X]\mathbb E[Y]]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]


We have


\mathbb E[(aX-b)(cY-d)]=\mathbb E[acXY-adX-bcY+bd]

=ac\mathbb E[XY]-ad\mathbb E[X]-bc\mathbb E[Y]+bd


\mathbb E[aX-b]=a\mathbb E[X]-b


\mathbb E[cY-d]=c\mathbb E[Y]-d


\mathbb E[aX-b]\mathbb E[cY-d]=ac\mathbb E[X]\mathbb E[Y]-ad\mathbb E[X]-bc\mathbb E[Y]+bd


Putting everything together, we find the covariance reduces to


\mathrm{Cov}(aX-b,cY-d)=ac(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])=ac\mathrm{Cov}(X,Y)


as desired.

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Hi!pls can u help me factorise this​
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Answer:

cant read

Step-by-step explanation:

:P

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Answer:

it is 1

Step-by-step explanation:

The power of 0 causes all numbers to negate to 1.
Anything to the power of 0 is 1


For example:

(10)^0 = 1

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2 years ago
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