Answer:
Step-by-step explanation:
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Answer:
a) distance covered by hare d1 = 8t
b) distance covered by tortoise d2 = 5t + 550
c) ∆d = 550 - 3t
Step-by-step explanation:
Given;
Speed of hare u = 8m/s
Speed of tortoise v = 5 m/s
Initial distance of tortoise d0 = 550 m
a) using the equation of motion;
distance covered = speed × time + initial distance
d = vt + d0
For hare;
d0 = 0
Substituting the values;
d1 = 8t + 0
d1 = 8t
b)using the equation of motion;
distance covered = speed × time + initial distance
d2 = vt + d0
For tortoise;
d0 = 550m
Substituting the values;
d2 = 5t + 550
d2 = 5t + 550 m
c) the number of meters the tortoise is ahead of the hare.
∆d = distance covered by tortoise - distance covered by hare
∆d = d2 - d1
Substituting the values;
∆d = (5t + 550) - 8t
∆d = 550 - 3t
Answer:
B
Step-by-step explanation:
yeah just pick B man lol
Answer:
-3 degrees C
Step-by-step explanation:
We need to subtract 3 degrees from Thursdays temperature
0 - 3
-3 degrees C
Answer: the probability that a randomly selected Canadian baby is a large baby is 0.19
Step-by-step explanation:
Since the birth weights of babies born in Canada is assumed to be normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = birth weights of babies
µ = mean weight
σ = standard deviation
From the information given,
µ = 3500 grams
σ = 560 grams
We want to find the probability or that a randomly selected Canadian baby is a large baby(weighs more than 4000 grams). It is expressed as
P(x > 4000) = 1 - P(x ≤ 4000)
For x = 4000,
z = (4000 - 3500)/560 = 0.89
Looking at the normal distribution table, the probability corresponding to the z score is 0.81
P(x > 4000) = 1 - 0.81 = 0.19
