Answer:
The mean age of the frequency distribution for the ages of the residents of a town is 43 years.
Step-by-step explanation:
We are given with the following frequency distribution below;
Age Frequency (f) X 
0 - 9 30 4.5 135
10 - 19 32 14.5 464
20 - 29 12 24.5 294
30 - 39 20 34.5 690
40 - 49 25 44.5 1112.5
50 - 59 53 54.5 2888.5
60 - 69 49 64.5 3160.5
70 - 79 13 74.5 968.5
80 - 89 <u> 8 </u> 84.5 <u> 676 </u>
Total <u> 242 </u> <u> 10389 </u>
Now, the mean of the frequency distribution is given by the following formula;
Mean = 
= 
= 42.9 ≈ 43 approx.
Hence, the mean age of the frequency distribution for the ages of the residents of a town is 43 years.
It's a half of an equilateral triangle because their angles measure 30° and 60° (an equilateral triangle has 3 angles of 60°)
So x=(2h/√3)/2=h√3=27√3≈15.6
Answer:
Below.
Step-by-step explanation:
Part A: 
Part B: Degree = 2, Classification = Quadratic
Part C: The polynomial is closed because it is multiplied. You CANNOT close under division.
Answer:
Step-by-step explanation:
Use the equation
89 < 75 + 0.7*M Subtract 75 from both sides
89 - 75 < 0.7M Combine
14 < 0.7 M Divide both sides by 0.7
14/0.7 < M
20 < M
If M goes over 20 then the flat rate is cheaper.
Answer:
0.010
Step-by-step explanation:
We solve the above question using z score formula
z = (x-μ)/σ, where
x is the raw score = 63 inches
μ is the population mean = 70 inches
σ is the population standard deviation = 3 inches
For x shorter than 63 inches = x < 63
Z score = x - μ/σ
= 63 - 70/3
= -2.33333
Probability value from Z-Table:
P(x<63) = 0.0098153
Approximately to the nearest thousandth = 0.010
Therefore, the probability that a randomly selected student will be shorter than 63 inches tall, to the nearest thousandth is 0.010.
