Two boats start their journey from the same point A and travel along directions AC and AD, as shown below: ABC is a right triang

The response is:

Therefore, option 1 is correct.

Hence, the final option is accurate.

Step-by-step breakdown:

Two functions are presented

. The domain represents the x values possible

. The range represents the y values achievable

. Thus, x can take any real number in functions f(x) and g(x)

. So, option 1 is correct.

For the range aspect

, f(x) can encompass any real number

. However, for g(x), the range can only include positive numbers

. Consequently, the last option is accurate, while all other options are incorrect.

The correct options include B "The sine function rises in the intervals (0°, 90°) and (270°, 360°)." E "Both sinusoidal functions reach a peak value of 1." and F "Both functions exhibit periodic behavior."

Answer:

50

Step-by-step clarification:

The equation that represents the total cost is in the format of a linear equation y = mx + c

Here, m signifies the slope of the line

c indicates the y-intercept, showing where the line intersects the y-axis

When the equation y = 150x + 50 is plotted, it will form a linear graph where the y-intercept corresponds to 50, as observed in the standard form of a linear equation.

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

There are 50 pairs of whole numbers that can add up to 99. To clarify, consider the combinations: 0 + 99, 1 + 98, 2 + 97, and so on, up to 49 + 50. This observation indicates that counting all combinations would result in 50 pairs of distinct whole numbers that sum to 99. I hope this response assists you!