Which measure is the largest? 1.2 km, 120,600 cm, 1,220,000 mm, 120 meters
1 answer:
The answer would be 1,220,000mm.
You can do this if you convert all the measures into one unit. Let us convert all into km.
The first one is already in km, so we do not need to covert it.
Let's start with converting 1,220,000mm to km.
There are 1, 000,000 mm in one km.
So there are 1.22km in 1,220,000mm.
Next, we have 120m. There are 1,000m in 1km.
There are 0.12km in 120m
Now you can see that 1,220,000mm is the longest.
You can do this if you convert all the measures into one unit. Let us convert all into km.
The first one is already in km, so we do not need to covert it.
Let's start with converting 1,220,000mm to km.
There are 1, 000,000 mm in one km.
So there are 1.22km in 1,220,000mm.
Next, we have 120m. There are 1,000m in 1km.
There are 0.12km in 120m
Now you can see that 1,220,000mm is the longest.
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Answer:
a) n(none) = 25
b) n(PE but not Bio) = 25
c) n(ENG but not both BIO and PE) = 55
d) n(students that did not take Eng or Bio) = 40
e) P( Students did not take exactly two subjects) = 0.65
Step-by-step explanation:
From the Venn diagram drawn:
a) Number of students that took none
n(Freshmen) = 135
n(all three) = 5
n (PE and Bio) = 10
n(PE and Eng) = 15
n(Bio and Eng) = 7
n (PE and Bio only) = 10 - 5 = 5
n(PE and Eng only) = 15 - 5 = 10
n(Bio and Eng only) = 7 - 5 = 2
n(PE only) = 35 - 5 - 5 - 10 = 15
n(Bio only) = 42 - 5 - 5 - 2 = 30
n(Eng only) = 60 - 10 - 5 -2 = 43
n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)
135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)
135 = 110 + n(none)
n(none) = 135 - 110
n(none) = 25
b)Number of students that too PE but not Bio
n(PE but not bio)= n(PE only) + n(PE and Eng only)
n(PE but not Bio) = 15 + 10
n(PE but not Bio) = 25
c) Number of students that took ENG but not both BIO and PE
n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10
n(ENG but not both BIO and PE) = 55
d) Number of students that did not take ENG or BIO
n( students that did not take Eng or Bio) = n(PE only) + n(none)
n(students that did not take Eng or Bio) = 15 + 25
n(students that did not take Eng or Bio) = 40
e) Probability that a randomly-chosen student from this group did not take exactly two subjects
n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)
n( Students that did not take exactly two subjects) = 15 + 30 + 43
n( Students that did not take exactly two subjects) = 88
P( Students did not take exactly two subjects) = 88/135
P( Students did not take exactly two subjects) = 0.65
Answer:
mean (μ) = 4.25
Step-by-step explanation:
Let p = probability of a defective computer components =
let q = probability of a non-defective computer components =
Given random sample n = 25
we will find mean value in binomial distribution
The mean of binomial distribution = np
here 'n' is sample size and 'p' is defective components
mean (μ) = 25 X 0.17 = 4.25
<u>Conclusion</u>:-
mean (μ) = 4.25