The approximations of the mean and the standard deviation are 233.3 and 229.82, respectively
<h3>How to determine the mean?</h3>
The table of values is given as:
Savings Lower Limit Upper Limit Frequency
0-199 0 199 345
200-399 200 399 97
400-599 400 599 52
600-799 600 799 21
800-999 800 999 9
1000-1199 1000 1199 8
1200-1399 1200 1399 3
Rewrite the table to include the class midpoint and the frequency
x f
99.5 345
299.5 97
499.5 52
699.5 21
899.5 9
1099.5 8
1299.5 3
The mean is calculated as:
So, we have:
Evaluate
Approximate
Hence, the approximation of the mean is 233.3
<h3>How to determine the standard deviation?</h3>
The standard deviation is calculated as:
So, we have:
Evaluate
Hence, the approximation of the standard deviation is 229.82
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By the remainder theorem it will be:-
4(3)^3 - 5(3)^2 + 2(3) + 11
= 80
the remainder is 80.
Answer: not sure
Step-by-step explanation:
After 5 s, the particle traverses a path subtended by an angle of
(3π/4 rad/s) (5 s) = 15π/4 rad
Now, 15π/4 rad = 15/8 * 2π rad, so the arc has a length equal to 15/8 of the circumference of the path:
15/8 * 2π * (3 cm) = 45π/4 cm ≈ 35.3 cm
3,000 is the smallest number and 300,000 is the largest number