Jay is letting her bread dough rise. After three hours, her bread dough is \dfrac{11}{5} 5 11 ​ start fraction, 11, divided by,

Answer:0.5 on the edge for 2020

Step-by-step explanation:

I completed the assessment!

<span>The system of equations that can determine if the commuter jet’s flight path crosses the restricted airspace is:

y = \frac{1}{4}(x - 10)^2 + 6  (i)
y = \frac{-27}{34}x - \frac{5}{17}  (ii)
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Here's why:
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The closed airspace boundary is defined by points (10, 6) and (12, 7).
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The commuter jet’s linear path runs from (-18, 14) to (16, -13).

Equation (i) describes the boundary since it fits both (10, 6) and (12, 7):

For (10, 6):
\frac{1}{4}(10-10)^2 + 6 = 6 (true)

For (12, 7):
\frac{1}{4}(12-10)^2 + 6 = 1 + 6 = 7 (true)

Equation (ii) represents the commuter jet’s path as it fits both (-18, 14) and (16, -13):

For (16, -13):
-13 = \frac{-27}{34} \times 16 - \frac{5}{17} = -13 (true)

For (-18, 14):
14 = \frac{-27}{34} \times (-18) - \frac{5}{17} = 14 (true)

By solving this system, we can confirm that the jet’s flight path intersects the closed airspace.

D1,..,d9 = 0,0,2,2,2,3,4,6,8 //there are 9 values, presented in ascending order
Q2 (median) = d5 = 2 //middle value
Q1 = (d2+d3) / 2 = (0+2)/2 = 1
(Q1 represents the median of d1,d2,d3,d4, but as there is no singular middle element among four, the average is computed)
Q3 = (d7+d8) / 2 = (4+6)/2 = 5
interquartile range = IQR = Q3 - Q1 = 5 -1 = 4
final answer: 4

Answer:

Tina's overall average speed for the journey is 56 km/h.

Step-by-step explanation:

Sean's travel time from Town A to Town B is 4 hours.

Sean's average speed is 70 km/h.

Using the motion equation, s = ut + 0.5 at², we can find:
Time, t = 4 hours.

Initial speed, u = 70 km/hr.

Acceleration, a = 0 m/s².

Replacing the variables:

      s = ut + 0.5 at²

      s = 70 x 4 + 0.5 x 0 x 4²

      s = 280 km

The distance from Town A to Town B is 280 km.

Tina's travel time exceeds Sean's by 1 hour.

Tina's time to travel from Town A to Town B is 4 + 1 = 5 hours.

Again applying the motion equation, s = ut + 0.5 at²:

Time, t = 5 hours.

Distance, s = 280 km.

Acceleration, a = 0 m/s².

Putting in values:

      s = ut + 0.5 at²

      280 = u x 5 + 0.5 x 0 x 5²

      u = 56 km/hr.

Tina's overall average speed is thus 56 km/h.

I couldn't write it out in detail, so the solution is 200. The work is visible in the image below if you need clarification. I hope this is useful to you!