Answer:
The confidence level is 0.98 and the significance is 
and 
and the critical value using the table is:
And replacing we got:
Step-by-step explanation:
For this case we have the following info given:
represent the population deviation
the sample size
represent the sample mean
We want to find the margin of error for the confidence interval for the population mean and we know that is given by:
The confidence level is 0.98 and the significance is and and the critical value using the table is:
And replacing we got:
Answer:
001,23,83,88,112,132,667
Step-by-step explanation:
Are you kidding me that you need help on this?
Step-by-step explanation:
1. 1st of all calculate the gradient
( - 3, 5) ( 2, 10)
Gradient = (10 - 5) / ( 2--3)
= 1
2. Then find the eq
Y = mx + c
Where m is the gradient
y= 1x + c
Now replace any 2 coordinates from above in the eq.
For ex I'm taking (2, 10)
Y = 1x + c
In the coordinate, x = 2 and y =10
By replacing this in the eq, I will find c
10 = 1(2) + c
2 + c = 10
c = 10 - 2
= 8
So the eq is y = x + 8 ⬅️
1 ) the formula used to calculate the simple interest is
I = Ci*n, f his formula you obtain n , that is the time in years n = I/ Ci, where c is the initial capital, i is the earned interest , and i is he rate of interest,
n = 120/600 (0,025) = 120 / 15 = 8 , at 8 years, you will obtain $ 120 of INTEREST
Answer:
A. Miguel has the greatest spread.
The range of the two persons can be determined by:
Adam's range = 106 -91
= 15
Miguel's range = 105 -86
= 19
B. Considering the middle 50% of the training time, the person with the least spread is Adam.
Adam - 103 105 104 106 100
Miguel - 88 86 89 93 105
Adam's 50% range = 106 - 100
= 6
Miguel's 50% range = 105 - 86
= 19
Adam has the least spread
C. Miguel is inconsistent with the time set for training compared to that of Adam.
The answers to parts 2a and 2b show that there is a wide variation in the time that Miguel spend during training, but a minimum variation in the time spent by Adam during training.
