A jogging track is 1/4 mile. Sara ran 14 laps.
2 answers:
Answer:
Sarah need to run more miles.
Step-by-step explanation:
In order to tell how much more she needs to run we need to find out how much she ran so far. Since we know that one lap is of a mile, and that she ran 14 laps we can easily find out how much she ran so far by multiplying
by 14. And so we get......
Now that we know how much she ran so far ( miles), in order to find out how much more she has to run we just have to substract the distance she ran so far from the total distance she wants to reach (in our case 5 miles) . After doing this we get the answer which is......
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Answer:
Part A
a. There is not sufficient sample evidence to conclude that the full-time placement rate is now less than 87% because the p-value is greater than 0.05
Part B
= 17.9·(x) + 1.6
Part C
The teen pregnancy rate in Florida is 6.54
Step-by-step explanation:
The percentage of graduates that find full-time employment in their fields within the first year of graduation = 87%
The significant level of the hypothesis test = 5% = 0.05
The p-value = 0.07
Given that the testing conditions are met, we have;
The p-value is larger than the significant level of 0.05, therefore, it cannot be concluded that there is significant difference between the two rates
Therefore;
There is not enough statistical evidence available to come to a conclusion that the placement rate of full-time students has reduced below the given 87% because from the statistics, the p-value is larger than 0.05.
Part B
Number of Pages, x; 7, 12, 4, 14, 25, 30
Cost, y; 128, 213, 75, 250, 446, 540
The regression analysis formula is;
Y = a + b·X
We have;
∴ b = (6×34602-92×1652)/(6×1930 - 92²) ≈ 17.852
a = (1652 - 17.852 × 92)/6 ≈ 1.6
∴ Y ≈ 17.9·(x) + 1.6
Part C
The given regression model is y = 4.3 + 1.4(x)
Therefore, at x = 1.6, we get;
y = 4.3 + 1.4 × 1.6 = 6.54
The suggests that the teen pregnancy rate in Florida is 6.54.