Answer:
the second answer Is the right one
Answer:
It is a good Estimator of the Population Mean because the distribution of the sample midrange is just same as the distribution of the random variable.
Step-by-step explanation: from the table,
Minimum value = 34
maximum values = 1084
The sample mid-range can be computed as:
(Min.value + max.value)/2
(34 + 1084)/2
Sample mid-range = 55
The sample midrange uses only a small portion of the data, but can be heavily affected by outliers.
It provides information about the skewness and heavy-tailedness of the distribution which is just same as the distribution of the random variable.
The nature of this distribution is not intuitive but the Central Limit in which it will approach a normal distribution for large sample size.
Answer:
155
Step-by-step explanation:
Using the recursive formula and a₁ = 1, then
a₂ = 3a₁ + 2 = 3(1) + 2 = 3 + 2 = 5
a₃ = 3a₂ + 2 = 3(5) + 2 = 15 + 2 = 17
a₄ = 3a₃ + 2 = 3(17) + 2 = 51 + 2 = 53
a₅ = 3a₄ + 2 = 3(51) + 2 = 153 + 2 = 155
Thus the 5 th term is 155
Answer:
2 : 5
Step-by-step explanation:
700 g : 1.75 kg
Since ratio should be between same units, so we'll convert either g into kg or kg into g. For convenience, kg will be converted to g
700 g : 1.75 kg × 1000
700 g : 1750 g
÷ 50 ÷50
14 : 35
÷7 ÷7
2 : 5
7 9/12 = 7.75
2 11/12 = 2.917
So we can round both of these up.
7.75 rounds to 8
2.917 rounds to 3
8 + 3 = 11
So the estimated sum is 11.
7 9/12 + 2 11/12 = 10 2/3
So the actual sum is 10 2/3.
The estimated sum is a pretty good estimate to the original number.