10 students per gender:
Boys: <span>four Xs over five and one X over zero, two, three, four, ten, and twelve.
5, 5, 5, 5, 0, 2, 3, 4, 10, 12 </span>→ 0, 2, 3, 4, 5, 5, 5, 5, 10, 12<span>
mean: 5.1
range: 12
</span><span>Girls: three Xs above eight, two Xs above three and four, and one X above two, six and seven.
8, 8, 8, 3, 3, 4, 4, 2, 6, 7 </span>→ 2, 3, 3, 4, 4, 6, 7, 8, 8, 8
<span>mean: 5.3
range: 6
The boys have the higher range while the girls have the higher mean value.
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Answer:
Step-by-step explanation:
You are to make 5 assemblies.
Each assembly requires the use of 1 Type A bolt.
To make the 5 assemblies, you need 5 Type A bolts.
The container of bolts has a total of 60 bolts.
The focus - Type A bolts - is 20 out of this 60.
The probability of obtaining a Type A bolt at all, is 20/60, which is = 1/3
(A) What is the probability of taking the exact number of Type A bolts you need for your 5 assemblies, if you randomly take 10 bolts from the container?
- The exact number of Type A bolts you need for the 5 assemblies is 5
1/3 × 5/10 = 5/30 = 1/6 = 0.167
(B) What is the probability of taking/having less than 5 Type A bolts out of the randomly selected 10 bolts? The solution is to sum up the following:
1/3 × 4/10 = 0.133
1/3 × 3/10 = 0.1
1/3 × 2/10 = 0.067
1/3 × 1/10 = 0.033
1/3 × 0/10 = 0
TOTAL = 0.333
Do you need decimal, exact form, or mixed number?
I believe if I'm understanding this right you can split this candy bar into 2 thirds. If you have 8 pieces left, you can get 2 groups of three out of 8.
I hope this helps! <3
