Answer:
C. Decreases the margin of error and hence increases the precision
Step-by-step explanation:
If we select a sample by Simple Random Sampling in a population of “infinite” size (a population so large that we do not know its size exactly), then the margin of error is given by
where
<em>Z = The Z-score corresponding to the confidence level
</em>
<em>S = The estimated standard deviation of the population
</em>
<em>n = the size of the sample.
</em>
As we can see, since n is in the denominator of the fraction and the numerator is kept constant, the larger the sample size the smaller the margin of error, so the correct choice is:
C. Decreases the margin of error and hence increases the precision
Answer:
55.734 mph
Step-by-step explanation:
Let's call the wind speed x and the time of flight t.
Then, we can write the following equation, knowing that the distance is equal speed times time of flight:
910 = (350+x) * t
660 = (350-x) * t
If we isolate t in both equations and compare its value, we have that:
910 / (350+x) = 660 / (350-x)
(350+x)/(350-x) = 910/660
350 + x = 1.3788 * (350-x)
350 + x = 482.58 - 1.3788x
2.3788x = 132.58
x = 55.734 mph
Answer:
It's 1/3
Step-by-step explanation:
2/6 is equivalent to 1/3, so you can rewrite the question as 2/3 - 1/3.
<span>You can probably just work it out.
You need non-negative integer solutions to p+5n+10d+25q = 82.
If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80.
So this is the same as n + 2d + 5q ≤ 16
So now you simply have to "crank out" the cases.
Case q=0 [ n + 2d ≤ 16 ]
Case (q=0,d=0) → n = 0 through 16 [17 possibilities]
Case (q=0,d=1) → n = 0 through 14 [15 possibilities]
...
Case (q=0,d=7) → n = 0 through 2 [3 possibilities]
Case (q=0,d=8) → n = 0 [1 possibility]
Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81
Case q=1 [ n + 2d ≤ 11 ]
Case (q=1,d=0) → n = 0 through 11 [12]
Case (q=1,d=1) → n = 0 through 9 [10]
...
Case (q=1,d=5) → n = 0 through 1 [2]
Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42
Case q=2 [ n + 2 ≤ 6 ]
Case (q=2,d=0) → n = 0 through 6 [7]
Case (q=2,d=1) → n = 0 through 4 [5]
Case (q=2,d=2) → n = 0 through 2 [3]
Case (q=2,d=3) → n = 0 [1]
Total from case q=2: 1 + 3 + 5 + 7 = 16
Case q=3 [ n + 2d ≤ 1 ]
Here d must be 0, so there is only the case:
Case (q=3,d=0) → n = 0 through 1 [2]
So the case q=3 only has 2.
Grand total: 2 + 16 + 42 + 81 = 141 </span>
