ALGEBRA Jake Duffy drove 12,568 miles

Answer:

At the α = 0.10 level, there is no substantial evidence indicating that the average vertical jump for students at this school differs from 15 inches.

Step-by-step explanation:

A hypothesis test is necessary to verify the assertion that the average vertical jump of students diverges from 15 inches.

The null and alternative hypotheses are:

The significance level is set at 0.10.

The sample mean recorded is 17, and the sample standard deviation is 5.37.

The degrees of freedom are calculated as df=(20-1)=19.

The t-statistic is:

The two-tailed P-value corresponding to t=1.67 is P=0.11132.

<pSince this P-value exceeds the significance level, the result is not significant. Therefore, the null hypothesis remains unchallenged.

At the α = 0.10 level, there is no compelling evidence that the average vertical jump of students at this school deviates from 15 inches.

The equivalent ratio is 3/4.

Question 1: (2.2, -1.4). Question 2: (1.33, 1). Providing a detailed analysis, the equations for the given lines are specified as (1) passing through points (0, 2.5) and (2.2, 1.4), and (2) through (0, -3) and (2.2, -1.4). We are tasked with locating a common solution or intersection of these equations. This leads to finding x = 2.2, and consequently y = -1.4. Therefore, the solution set is (2.2, -1.4). For question 2, the equations yield a solution of (1.33, 1).

<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.