It shows that the values of the mileages are close to 24 mpg. Hence statistics refes 24 as the mean.
<h3>Mean of a data.</h3>
Mean is one of the measure of dispersion and is the avearage of a set of data.
According to the question, the magazine reports that the most common gas mileage was 24 mpg. This shows that reports have published several gas mileages for cars and truck and the average of this mileage is 24mpg
It shows that the values of the mileages are close to 24 mpg. Hence statistics refes 24 as the mean.
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Answer:
Probability = 0.12025
Step-by-step explanation:
P (Am) = 1/50 = 0.02 {Magazine ad}
P (At) = 1/8 = 0.125 {Television ad}
P (Am ∩ At ) = 1/100 = 0.01 {Both ads}
P (Am U At) = P (Am) + P (At) - P (Am ∩ At )
= 0.02 + 0.125 - 0.01
P (Am U At) = 0.135 {Person sees either ad}
P (Am' ∩ At') = 1 - P (Am U At)
P (Am' ∩ At') = 1 - 0.135 = 0.865 {Person sees none ad}
Prob (Purchase) = Prob (Purchase with ad) + Prob (purchase without ad)
P (P/ A) = 1/4 = 0.25 , P (P / A') = 1/10 = 0.1
P (P) = (0.25) (0.135) + (0.1) (0.865)
= 0.03375 + 0.0865
0.12025
Answer:
y= 2x + 10
Step-by-step explanation:
The variable depends on the $2 so, the green box should be 2
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
