Three friends went running. The table below shows their last 3 lap times.

Answer:

The exact population of Vermont in 2008 could have been around 618,000

Step-by-step explanation:

* Here's how rounding to the nearest ten thousand works:

- Numbers ending with the last four digits between 0001 and 4999 are rounded down to the nearest lower multiple of ten thousand

- Example: 83,525 rounds down to 80,000.

- If the last four digits are 5000 or above, round up to the next higher ten thousand

- Example: 58,988 rounds up to 60,000

* Applying this rule to the problem given:

Since the rounded population is 620,000 to the nearest ten thousand, the actual population could be any value with the last four digits from 0001 to 4999 (like 618,000) or from 5000 upwards (like 624,000).

Therefore, 618,000 might represent the actual population of Vermont in 2008

The y-intercept is now 12.5, and the slope has increased to 0.5625. After the policy adjustment, an additional $5 was included, followed by multiplying the total amount by 1.25.

C. -31m⁴n - 8m²Step-by-step explanation:Given:(9mn - 19m⁴n) - (8m² + 12m⁴n + 9mn)Required:Identify an equivalent expression for itSolution:Distributing the negative sign across the parentheses results in:9mn - 19m⁴n - 8m² - 12m⁴n - 9mnNext, we combine like terms:9mn - 9mn - 19m⁴n - 12m⁴n - 8m²This simplifies to -31m⁴n - 8m²Thus, -31m⁴n - 8m² is the equivalent expression for (9mn - 19m⁴n) - (8m² + 12m⁴n + 9mn).

Answer:

A), B), and C) are clarified below.

Step-by-step explanation:

The inquiry involves using binary digits, employing probabilities that are equal for both conditions, by applying a random test pattern, where the formula is derived from p = q.

Simplifying gives us

P[k] = nCk / 2^n

A. Probability of all bits being 1s

16c16/2^16 = 1/65536

B. Probability of all bits being 0s

16c0/2^16 = 1/65536

C. The probability of having exactly 8 bits as 1s and the other 8 as 0s

16c8/2^16 = 12870/65536 => 0.1963 ≈ 19.63%

Answer:

118.2°

Step-by-step explanation:

Dos líneas paralelas x e y son cortadas por la transversal w (ver el diagrama adjunto).

Se forman 8 ángulos (denominados 1, 2, 3, 4, 5, 6, 7 y 8).

Los ángulos 1 y 6 son ángulos del mismo lado cuando dos líneas paralelas x e y son cortadas por la transversal w.

Dos ángulos del mismo lado son suplementarios (suman 180°). Esto significa

Dado por lo que