Answer:
No
Step-by-step explanation:
the mean is 16 (add 4+5+5+8, then divide answer by 4) and the mode is 5 (occurs most in data)
Answer:
(5.216) mean = 61.71
(5.217) standard deviation = 8.88
(5.218) P(X>=60) = 0.9238
(5.129) L = 79, U = 69
(5.130) P(X> or = U) = 0.7058
Step-by-step explanation:
The table of the statistic is set up as shown in attachment.
(5.216) mean = summation of all X ÷ no of data.
mean = 432/7 = 61.71 birds
(5.217)Standard deviation = √ sum of the absolute value of difference of X from mean ÷ number of data
S = √ /X - mean/ ÷ 7
= √551.428/7
S = 8.88
(5.218) P (X> or = 60)
= P(Z> or =60 - 61.71/8.8 )
= P(Z>or= - 0.192)
= 1 - P(Z< or = 0.192)
= 1- 0.0762
= 0.9238
(5.219)the 15th percentile=15/100 × 7
15th percentile = 1.05
The value is the number in the first position and that is 79,
L= 79
85th percentile = 85/100 × 7 = 5.95
The value is the number in the 6th position, and that is 69
U = 69
5.130) P(X>or = 60)
= P(Z>or= 69 - 61.71/8.8)
= P(Z> or = 0.8208)
= 1 - P(Z< or = 0.8209)
= 1 - 0.2942
= 0.7058
Answer:
Bus speed: 22 mph; Trolley speed: 14 mph
Step-by-step explanation:
Let rb represent the speed of the bus and rt the speed of the trolley. Then rb = rt + 8 (mph).
Recall that distance = rate times time. Two different distances and two different speeds are involved here, but only one time.
Thus,
60 mi 44 mi
--------------------- = time (same in both cases) = -------------
rt + 8 mi/hr rt
Cross-multiplying, we get 44rt + 8(44) = 60 rt, or
352 = 16rt.
Solving this for rt, we get rt = speed of bus = 352/16 = 22 mph
The speed at which the bus travels is 22 mph and that at which the trolley travels is (22 mph - 8 mph) = 14 mph
