Answer:
Given:
Body mass index values:
17.7
29.4
19.2
27.5
33.5
25.6
22.1
44.9
26.5
18.3
22.4
32.4
24.9
28.6
37.7
26.1
21.8
21.2
30.7
21.4
Constructing a frequency distribution beginning with a lower class limit of 15.0 and use a class width of 6.0.
we have:
Body Mass Index____ Frequency
15.0 - 20.9__________3( values of 17.7, 18.3, & 19.2 are within this range)
21.0 to 26.9__________8 values are within this range)
27.0 - 32.9____________ 5 values
33.0 - 38.9____________ 2 values
39.0 - 44.9 _____________2 values
The frequency distribution is not a normal distribution. Here, although the frequencies start from the lowest, increases afterwards and then a decrease is recorded again, it is not normally distributed because it is not symmetric.
Answer:
the probability that randomly selected applicants over 10 years of experience is 0.6791
Step-by-step explanation:
The computation of the probability that randomly selected applicants over 10 years of experience is as follows:
Total would be
= 187 have 10 + years experience
Now
P(graduate | 10+) = (graduate and 10+ years experience) ÷ (10 + years of experience)
= 127 ÷ 187
= 0.6791
Hence, the probability that randomly selected applicants over 10 years of experience is 0.6791
Let x = 63.63...(by repeating i assume you mean the 63 is repeated as 63.636363...)
therefore 100x = 6363.63..
100x-x =6300 (the recurring bit has been cancelled out which is what you should always aim to do)
99x=6300
x = 6300/99
cancel down
x = 700/11
Answer:
H(t) = 15 -6sin(2.5π(t -0.5))
Step-by-step explanation:
For midline M, amplitude A, period T and time t0 at which the function is decreasing from the midline, the function can be written as ...
H(t) = M -Asin(2π/T(t -t0))
Using the given values of M=15, A=6, T=0.8 and t0 = 0.5, the equation is ...
H(t) = 15 -6sin(2.5π(t -0.5))
= 10