Describe how to find the number of seats in the middle and lower levels of the stadium when solving for the variable only gives

(a) The multiplicative inverse of 1234 (mod 4321) is x so that 1234*x ≡ 1 (mod 4321). We can apply Euclid's algorithm:

4321 = 1234 * 3 + 619

1234 = 619 * 1 + 615

619 = 615 * 1 + 4

615 = 4 * 153 + 3

4 = 3 * 1 + 1

Now we will express 1 as a linear combination of 4321 and 1234:

1 = 4 - 3

1 = 4 - (615 - 4 * 153) = 4 * 154 - 615

1 = 619 * 154 - 155 * (1234 - 619) = 619 * 309 - 155 * 1234

1 = (4321 - 1234 * 3) * 309 - 155 * 1234 = 4321 * 309 - 1082 * 1234

This reduces to

1 ≡ -1082 * 1234 (mod 4321)

Thus, the inverse is

-1082 ≡ 3239 (mod 4321)

(b) Since both 24140 and 40902 are even, their GCD cannot equal 1, indicating no inverse exists.

The options you provided are unclear to me, so I will respond in general terms: to determine if a point (ordered pair) lies on a line, you need to substitute the x-value from that pair into the line's equation and check if the resulting y-value matches the y-value of the ordered pair. For example, if your line is y = 4/3x + 1/3, we can check if (0, 0) and (2, 3) fit this line. We find that y = 4/3·0 + 1/3 gives us 1/3, which does not equal 0, indicating (0, 0) is not on the line. For (2, 3), substituting yields y = 4/3·2 + 1/3 = 3, meaning (2, 3) is on the line.

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:

The result is 26.4

Step-by-step explanation:

By taking 51.82 and subtracting 26.37, you arrive at 26.45, and when rounding this value to one decimal place, it becomes 26.4.

<span>The graph will shift 5 units to the right and 1 unit upwards, forming a parabola that opens up with its vertex positioned at (5, 1).

Explanation:
The subtraction of 5 from x prior to squaring indicates a horizontal movement of 5 units to the right.

The addition of 1 signifies a vertical shift of 1 unit up.

This transformation follows the vertex form of a parabola, y=a(x-h)^2 + k, where (h, k) represents the vertex. In this case, h is 5 and k is 1, placing the vertex at (5, 1).</span>