Using the information given above, the sampling distribution of the sample proportion of 100-ohm gold-band is 2.
- <em>Sampling distribution of proportion, P = 2% = 0.02 </em>
- <em>Sample size, n = 100</em>
<u>The sampling distribution of the sample proportion can be calculated thus</u>:
- <em>Distribution of sample proportion = np</em>
Distribution of sample proportion = (100 × 0.02) = 2
Therefore, there is a probability that only 2 of the samples will have resistances exceeding 105 ohms.
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<span>circumference = 60*PI cm
If an arc is 140 degrees of the circle then its length =
</span>
<span>60*PI cm * (140 / 360)
arc length = 23.33333 cm
</span>
Answer:
The one with arrows are the answers
->Line segment E B is bisected by Line segment D F .
->A is the midpoint of Line segment F C .
Line segment F C bisects Line segment D B.
->Line segment E B is a segment bisector.
->FA = One-halfFC.
Line segment D A is congruent to Line segment A B .
Step-by-step explanation:
I did it on edge and got it right
Answer:
500 total pages. 190 pages with no pictures.
Step-by-step explanation:
310 is 0.62 of the total pages. 310 divided bye 0.62 is 500.
500 minus 310 is 190 pages with no pictures.
Hope this helps!
Answer: A committee of 5 students can be chosen from a student council of 30 students in 142506 ways.
No , the order in which the members of the committee are chosen is not important.
Step-by-step explanation:
Given : The total number of students in the council = 30
The number of students needed to be chosen = 5
The order in which the members of the committee are chosen does not matter.
So we Combinations (If order matters then we use permutations.)
The number of combinations of to select r things of n things = 
So the number of ways a committee of 5 students can be chosen from a student council of 30 students=
Therefore , a committee of 5 students can be chosen from a student council of 30 students in 142506 ways.
