Answer:
Follows are the solution to this question:
Step-by-step explanation:
In the given question some of the data is missing so, its correct question is defined in the attached file please find it.
Let
A is quality score of A
B is quality score of B
C is quality score of C
![\to P[A] =0.55\\\\\to P[B] =0.28\\\\\to P[C] =0.17\\](https://tex.z-dn.net/?f=%5Cto%20P%5BA%5D%20%3D0.55%5C%5C%5C%5C%5Cto%20P%5BB%5D%20%3D0.28%5C%5C%5C%5C%5Cto%20P%5BC%5D%20%3D0.17%5C%5C)
Let F is a value of the content so, the value is:
![\to P[\frac{F}{A}] =0.15\\\\\to P[\frac{F}{B}] =0.12\\\\\to P[\frac{F}{C}] =0.14\\](https://tex.z-dn.net/?f=%5Cto%20P%5B%5Cfrac%7BF%7D%7BA%7D%5D%20%3D0.15%5C%5C%5C%5C%5Cto%20P%5B%5Cfrac%7BF%7D%7BB%7D%5D%20%3D0.12%5C%5C%5C%5C%5Cto%20P%5B%5Cfrac%7BF%7D%7BC%7D%5D%20%3D0.14%5C%5C)
Now, we calculate the tooling value:
![\to p[\frac{C}{F}]](https://tex.z-dn.net/?f=%5Cto%20p%5B%5Cfrac%7BC%7D%7BF%7D%5D)
using the baues therom:
![\to p[\frac{C}{F}] = \frac{p[C] \times p[\frac{F}{C} ]}{p[A] \times p[\frac{F}{A}] + p[B] \times p[\frac{F}{B}]+p[C] \times p[\frac{F}{C}] }](https://tex.z-dn.net/?f=%5Cto%20p%5B%5Cfrac%7BC%7D%7BF%7D%5D%20%3D%20%20%5Cfrac%7Bp%5BC%5D%20%5Ctimes%20p%5B%5Cfrac%7BF%7D%7BC%7D%20%5D%7D%7Bp%5BA%5D%20%5Ctimes%20p%5B%5Cfrac%7BF%7D%7BA%7D%5D%20%2B%20p%5BB%5D%20%5Ctimes%20p%5B%5Cfrac%7BF%7D%7BB%7D%5D%2Bp%5BC%5D%20%5Ctimes%20p%5B%5Cfrac%7BF%7D%7BC%7D%5D%20%7D)
The functions are illustrations of composite functions.
<em>The soil temperature at 2:00pm is 67</em>
The given parameters are:
---- the function for sun intensity
-- the function for temperature
At 2:00pm, the value of h (number of hours) is:
Substitute 8 for h in , to calculate the sun intensity
Substitute 8/9 for I in , to calculate the temperature of the soil
Approximate
Hence, the soil temperature at 2:00pm is 67
Read more about composite functions at:
brainly.com/question/20379727
Answer:
a.) one sample t test
b.) H0 : μ = 59.3
c.) H1 : μ > 59.3
d.) μ = 59.3 ; σ = 39.84
e.) xbar = 79.4 ; s = 61.36
Test statistic = 3.16
Step-by-step explanation:
Given the sample data:
49.00 49.00 49.00 49.00 49.00 63.00 63.00 63.00 63.00 63.00 199.00 199.00 199.00 199.00 199.00 38.00 38.00 38.00 38.00 38.00 48.00 48.00 48.00 48.00 48.00 49.00 63.00 199.00 38.00 48.00
Sample size, n = 30
Using calculator :
xbar from the data above = 79.4
Standard deviation = 61.359
H0 : μ = 59.3
H1 : μ > 59.3
Test statistic :
(Xbar - μ) ÷ (σ/sqrt(n)
σ = 34.83
(79.4 - 59.3) ÷ (34.83/sqrt(30))
20.1 ÷ 6.359
Test statistic = 3.16
