A commander wants to arrange his soldiers in such a way that the number of row is equal to the number of columns. If the total n

Answer:

Given that the frog jumps every 10 seconds

   (using digits from a random number table)

• It requires 7 jumps with 2 in the reverse direction (either left or right) for the frog to get off the board in 60 seconds.
• Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
• Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.

Step-by-step explanation:

A frog positioned right at the center of a 5ft long board is 2.5 ft away from either edge.

Every 10 seconds, the frog jumps left or right.

If the frog's jumps are LLRLRL, it will remain on the board at the leftmost square.

If it jumps as LLRLL, it will jump off the board after fifty seconds.

Given that the frog jumps every 10 seconds

   (using digits from a random number table)

• It requires 7 jumps with 2 in reverse direction (either left or right) for the frog to get off the board in 60 seconds.
• Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
• Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.

Answer:

∠ R 90°

   

Explicación paso a paso:

Dado que en el triángulo RST

Ahora, según la condición, un ángulo es mayor que la suma de los otros dos ángulos.

En un triángulo, la suma de los tres ángulos es 180°

Por lo tanto, si un ángulo mide  90°, la suma de los otros dos debe ser igual a 90°

Y si uno de los ángulos es de 90°, solo los otros dos pueden ser de 45° cada uno.

Aquí suponiendo que el ángulo s = ángulo T = 30°, entonces con esta condición el ángulo r sería de 120°, que es mayor que 90°.

De lo anterior, se concluye que

Para,

∵ ∠ R = 120°, por lo tanto es mayor que 90°.

Es decir, ∠ R 90°

  Respuesta

In order to determine this probability, we calculate using this difference: To obtain these probabilities, it’s possible to utilize normal standard distribution tables, a calculator, or software like Excel. The accompanying figure displays the results achieved. Here’s a detailed breakdown of the steps: Relevant concepts include the normal distribution, which describes a probability distribution that is symmetric regarding the mean, demonstrating that occurrences close to the mean are more likely than those farther away. The Z-score represents a statistical measure illustrating how far a value is from the average of a set, expressed in standard deviations. For our analysis, let X denote the random variable representing weights in a population, with its distribution characterized by: We’re specifically interested in this probability. The most effective approach to address this issue is through the standard normal distribution and the Z-score calculation, expressed as: Applying this formula to our probability provides the following: This allows us to calculate this probability with the provided difference: We use standard distribution tables, a calculator, or Excel for determining these probabilities. The graph illustrates the resulting outcome.

Answer:

a. Alpha equals 3.014 while beta equals 12.442

b. The likelihood that the data transfer duration surpasses 50ms is 0.238

c. The chance that data transfer time falls between 50 and 75 ms is 0.176

Step-by-step explanation:

a. Given the data, the mean and standard deviation for the random variable X are 37.5 ms and 21.6, respectively.

Thus, E(X)=37.5 and V(X)=(21.6)∧2  

To find alpha, we need to apply the formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To determine beta, the following formula is employed:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. With E(X)=37.5 and V(X)=(21.6)∧2,  

Hence, P(X>50)=1−P(X≤50)

To find the probability of data transfer time exceeding 50ms, we use the formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The chance of data transfer time exceeding 50ms is 0.238

c. With E(X)=37.5 and V(X)=(21.6)∧2,  

Thus, P(50<X<75)=P(X<75)−P(X<50)  

To find the probability that data transfer time is between 50 and 75 ms, we apply the formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

​

The probability that data transfer time falls between 50 and 75 ms is 0.176

Answer:

 -1.7x -3

Step-by-step explanation:

−4.2x+3 +2.5x−6

I prefer to align them one above the other.

−4.2x+3

 2.5x−6

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 -1.7x -3