Out of 100 surveyed individuals, 35 work out in the morning, 45 in the afternoon, and 20 at night.
Jim is accurate.....
The ratio of morning exercisers to the whole group is 35/100.
The ratio of afternoon exercisers is 45/100.
The ratio for night exercisers is 20/100.
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.
= -4/2 Step-by-step explanation: xp=x1 +k (x2-x1) = -3+1/3 (2- -3) = -3 + 1/3 (5/1) = -3 + 5/3 = -9/3 +5/3 = -4/2 Sorry, it doesn't look the best:)
Additional 300 grams of flour will be required.
The function is applicable within the segments of x:
(-∞, -1) and [-1, 7), meaning it is valid for x < 7.
Importantly,
the function cannot be evaluated at x = -1 in the left part of the linear graph, while it is valid at x = -1 in the right segment of the same line. Additionally, the function is not defined at x = 7 or any value above it.
Conclusion: x < 7.