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Answer:

Compound interest is that which, once generated, is added to the capital, in order to expand the calculation base on which, in turn, new interests will be generated. Thus, in short, if the interest is, for example, 5 percent, said interest will be added to the initial capital, with which that 5% generated will be increasing, since it will be calculated on an increasingly large basis.

Compound interest is very important for financial investments, as it maximizes the results of said investments, even more so when compared to simple interest, in which the interest generated is not added to the initial capital.

Answer:

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Coin chosen from box B is red.

Event B: Blue poker chip transferred.

Probability of choosing a red coin:

7/10 of 4/9(red coin from box A)

6/10 of 5/9(blue coin from box A). So

[tex]P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444[/tex]

Blue chip transferred, red coin chosen:

6/10 of 5/9. So

[tex]P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333[/tex]

What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172[/tex]

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

Answer:

Step-by-step explanation:

Percentage decrease = Difference/Original value

So,

Percentage decrease

[tex] = \frac{2}{20} \times 100 \\ = 2 \times 5 \\ = 10[/tex]

So there fore the percentage decrease is 10%

Answer:

Draw a line going from 360 on the vertical line to 30 on the horizontal line.

Step-by-step explanation:

12 meters a minute. 30 minutes. 360 meters in total.

Answer:

I believe the correct answer is C

Answer:

It should be A) 54

Step-by-step explanation:

32 + 14 + 8 = 54

Answer:

Step-by-step explanation:

Use SOH CAH TOA to recall how the trig functions fit on a triangle

SOH: Sin(Ф)= Opp / Hyp

CAH: Cos(Ф)= Adj / Hyp

TOA: Tan(Ф) = Opp / Adj

Tan(65) = Opp / 10

10* Tan(65) = Opp

21.445 = Opp

so the height of the tree is approx  21.4 feet ( to the nearest 10th )

The medians are about the same.

Option C is the correct answer.

It is the middle value of the given set of numbers after arranging the given set of numbers in order.

We have,

Movie theater 1.

From the box plot given,

The median is 995.

Movie theater 2.

From the box plot given,

The median is 995.

Thus,

The median in both the box plot is the same.

Learn more about the median here:

https://brainly.com/question/28060453

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Answer:

Hello good afternoon

Step-by-step explanation:

you have to put this link

Answer:

you would do 4.5 * 250 and then you would get an answer of 1125

Step-by-step explanation:

hope this helps have a great rest of your day/night

Answer:

40 feet

Step-by-step explanation:

Answer:

x = 15

Step-by-step explanation:

15 * 15 = 225

Answer:

X = 15

Step-by-step explanation:

This is because if you square 15, you get 225.

Answer:

A=-7

Step-by-step explanation:

7=35+A4

4A=-28

A=-7

   Factoring  x2-14x+11 

The first term is,  x2  its coefficient is  1 .

The middle term is,  -14x  its coefficient is  -14 .

The last term, "the constant", is  +11 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 11 = 11 

Step-2 : Find two factors of  11  whose sum equals the coefficient of the middle term, which is   -14 .

     -11   +   -1   =   -12     -1   +   -11   =   -12     1   +   11   =   12     11   +   1   =   12

Observation : No two such factors can be found

Answer:

=x2+−25+0

=x2+−25+0

=(x2)+(−25+0)

=x^2+−25

Aqui esta la respuesta y el procedimiento

Answer:

D) 20º

Step-by-step explanation:

if one angle is 70º, and the other is 90º, that means that the first two angles equal 160º. since all three interior angles of a triangle have to add up to 180º, the third angle must be 20º.

Answer:

They add up to 720 degrees

Answer:

360 I think. Hope i helped! :)

Answer:

65

Step-by-step explanation:

( (23 * 13) / 2 ) - ( (13 * 13) / 2 ) = ( 299 / 2 ) - ( 169 / 2 ) = 149.5 - 84.5 = 65

Numbers are slightly unclear but I think this is right.

Answer:

3. D 4. B

Step-by-step explanation:

3.  e^2 = g^2 + f^2, answer is d

4. Let a = 3, b = 4, c = 5

2c = 10

2b = 8

2a = 6

10^2 = 8^2 + 6^2

answer is b

3. The Pythagorean theorem states the hypotenuse squared is equal to each side squared added to each other.

The answer is D. E^2 = f^2 + g^2

4.

To keep the proportion you could multiply the sides by the same numbers . The answer would be 2A, 2b, 2c

Answer:

D

Step-by-step explanation:

Answer:

1,695.6 in. 3

Step-by-step explanation:

Answer: 12.35

Step-by-step explanation: Since we know that there is a total of 24.7 grams in both piles, we can find how much salt is in one by dividing 24.7 by 2.

The amount of salt in each pile will be the half amount of 24.7 grams thus 12.35 grams will be in each pile.

Arithmetic operators are four basic mathematical operations in which summation, subtraction, division, and multiplication involve.,

Division = divide any two numbers or variables called division.

For example 4/8

As per the given,

Amount of salt available = 24.7 grams.

It needed to divide into two piles.

In one piles = 24.7/2 = 12.35 grams

Hence "The amount of salt in each pile will be the half amount of 24.7 grams thus 12.35 grams will be in each pile".

To learn more about the arithmetic operators,

brainly.com/question/25834626

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Answer:

[tex]\begin{array}{c|c|c||c}X & Y & Z & (X \to Y) \lor (Z \to \lnot X) \\ \cline{1-4} \rm T & \rm T & \rm T & \rm T\\ \rm T & \rm T & \rm F & \rm T \\ \rm T & \rm F & \rm T & \rm F \\ \rm T & \rm F & \rm F & \rm T \\ \cline{1-4} \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T \\ \rm F & \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T\end{array}[/tex].

[tex]\begin{array}{c|c|c||c}X & Y & Z & (X \to Y) \lor (Z \to \lnot X) \\ \hline \rm T & \rm T & \rm T & \rm T\\ \rm T & \rm T & \rm F & \rm T \\ \rm T & \rm F & \rm T & \rm F \\ \rm T & \rm F & \rm F & \rm T \\ \hline \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T \\ \rm F & \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T\end{array}[/tex].

Step-by-step explanation:

Let [tex]A[/tex] denote a Boolean variable.

The negation of [tex]A[/tex] ([tex]\lnot A[/tex]) is false if when [tex]\!A[/tex] is true, and true when [tex]A\![/tex] is false. In a truth table:

[tex]\begin{array}{c||c} A & \lnot A \\ \cline{1-2} \rm T & \rm F \\ \rm F & \rm T\end{array}[/tex].

[tex]\begin{array}{c||c} A & \lnot A \\ \hline \rm T & \rm F \\ \rm F & \rm T\end{array}[/tex].

Let [tex]B[/tex] denote another Boolean variable. The material implication "[tex]A[/tex] implies [tex]\!B[/tex]" ([tex]A \to B[/tex]) is true unless [tex]B\![/tex] is false when [tex]A\![/tex] is true.  

[tex]\begin{array}{c|c||c} A & B & A \to B \\ \cline{1-3} \rm T & \rm T & \rm T \\ \rm T & \rm F & \rm F \\ \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm T\end{array}[/tex].

[tex]\begin{array}{c|c||c} A & B & A \to B \\ \hline \rm T & \rm T & \rm T \\ \rm T & \rm F & \rm F \\ \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm T\end{array}[/tex]

The logical or "[tex]A[/tex] or [tex]B[/tex]" is true when either [tex]A\![/tex] or [tex]B\![/tex] is true (and also when both are true.)

[tex]\begin{array}{c|c||c} A & B & A \lor B \\ \cline{1-3} \rm T & \rm T & \rm T \\ \rm T & \rm F & \rm T \\ \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm F\end{array}[/tex]

[tex]\begin{array}{c|c||c} A & B & A \to B \\ \hline \rm T & \rm T & \rm T \\ \rm T & \rm F & \rm F \\ \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm T\end{array}[/tex].

Start by finding the value of [tex]\lnot X[/tex], [tex](X \to Y)[/tex], and [tex](Z \to \lnot X)[/tex] for each of the [tex]2^3 = 8[/tex] possible combinations of [tex]X[/tex], [tex]Y[/tex], and [tex]Z[/tex].

[tex]\begin{array}{c|c|c||c||c|c}X & Y & Z & \lnot X & (X \to Y) & (Z \to \lnot X) \\ \cline{1-6} \rm T & \rm T & \rm T & \rm F & \rm T & \rm F\\ \rm T & \rm T & \rm F & \rm F & \rm T & \rm T \\ \rm T & \rm F & \rm T & \rm F & \rm F & \rm F \\ \rm T & \rm F & \rm F & \rm F & \rm F & \rm T \\ \cline{1-6} \rm F & \rm T & \rm T & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm F & \rm T & \rm T & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T & \rm T & \rm T \end{array}[/tex].

[tex]\begin{array}{c|c|c||c||c|c}X & Y & Z & \lnot X & (X \to Y) & (Z \to \lnot X) \\ \hline \rm T & \rm T & \rm T & \rm F & \rm T & \rm F\\ \rm T & \rm T & \rm F & \rm F & \rm T & \rm T \\ \rm T & \rm F & \rm T & \rm F & \rm F & \rm F \\ \rm T & \rm F & \rm F & \rm F & \rm F & \rm T \\ \hline \rm F & \rm T & \rm T & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm F & \rm T & \rm T & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T & \rm T & \rm T \end{array}[/tex].

The value of [tex](X \to Y) \lor (Z \to \lnot X)[/tex] is true whenever either [tex](X \to Y)[/tex] or [tex](Z \to \lnot X)[/tex] is true (or both.) The combination [tex]X = \rm T[/tex], [tex]Y = \rm F[/tex], and [tex]Z = \rm T[/tex] is the only one among the eight where neither [tex](X \to Y)\![/tex] nor [tex](Z \to \lnot X)\![/tex] is true. [tex](X \to Y) \lor (Z \to \lnot X)\![/tex] would evaluate to true for all other combinations.

Hence, the truth table would be:

[tex]\begin{array}{c|c|c||c}X & Y & Z & (X \to Y) \lor (Z \to \lnot X) \\ \cline{1-4} \rm T & \rm T & \rm T & \rm T\\ \rm T & \rm T & \rm F & \rm T \\ \rm T & \rm F & \rm T & \rm F \\ \rm T & \rm F & \rm F & \rm T \\ \cline{1-4} \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T \\ \rm F & \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T\end{array}[/tex].

[tex]\begin{array}{c|c|c||c}X & Y & Z & (X \to Y) \lor (Z \to \lnot X) \\ \hline \rm T & \rm T & \rm T & \rm T\\ \rm T & \rm T & \rm F & \rm T \\ \rm T & \rm F & \rm T & \rm F \\ \rm T & \rm F & \rm F & \rm T \\ \hline \rm F & \rm T & \rm T & \rm T \\ \rm F & \rm T & \rm F & \rm T \\ \rm F & \rm F & \rm T & \rm T \\ \rm F & \rm F & \rm F & \rm T\end{array}[/tex].