The answer is 5. |-12+7| =|-5|=5
Part A
(85 km)/(51 min) = (x km)/(60 min)
85/51 = x/60
85*60 = 51x
5100 = 51x
51x = 5100
x = 5100/51
x = 100
<h3>Answer: 100 km</h3>
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Part B
510 m = 0.510 km
(85 km)/(51 min) = (0.510 km)/(x min)
85/51 = 0.510/x
85x = 51*0.510
85x = 26.01
x = 26.01/85
x = 0.306
This represents the number of minutes. Multiply by 60 to get the number of seconds
0.306 min = 0.306*60 = 18.36
<h3>Answer: 18.36 seconds</h3>
Answer:
And we are 9% confidence that the true mean for the difference of the population means is given by:
Step-by-step explanation:
For this problem we have the following data given:
represent the sample mean for one of the departments
represent the sample mean for the other department
represent the sample size for the first group
represent the sample size for the second group
represent the deviation for the first group
represent the deviation for the second group
Confidence interval
The confidence interval for the difference in the true means is given by:
The confidence given is 95% or 9.5, then the significance level is and . The degrees of freedom are given by:
And the critical value for this case is
And replacing we got:
And we are 9% confidence that the true mean for the difference of the population means is given by:
Answer:
24. a) ⅞ b) 1⅘
17. 36
Step-by-step explanation:
24.
a) ½ × ¾ + 3/2 × ⅓
3/8 + 1/2
(3 + 4×1)/8
7/8
b) 1½ ÷ (1½ - ⅔)
3/2 ÷ (3/2 - 2/3)
3/2 ÷ (3×3 - 2×2)/6
3/2 ÷ 5/6
3/2 × 6/5
18/10
1 8/10
1⅘
17. 1 day = 24 hours
1.5 days = 1.5× 24= 36 hours
Team A has more spread out times, but they have the fastest times in their lower quartile.
Team B has more consistent times, but their times in the lower quartile are slower than Team A's. Their upper quartile times are faster than Team A's.
Both teams observed the same fastest and lowest times.
Both teams' medians are close to each other, but Team B has a slightly faster median.