Answer:
The 93% confidence interval for the equatorial radius of Jupiter is between 71484 km and 71500 km.
Step-by-step explanation:
Sample size of 30 or larger, so we can use the normal distribution to find the confidence interval.
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so 
Now, find M as such
In which 
is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 71492 - 8 = 71484 km.
The upper end of the interval is the sample mean added to M. So it is 71492 + 8 = 71500 km.
The 93% confidence interval for the equatorial radius of Jupiter is between 71484 km and 71500 km.
The first thing we must do for this case is to define variables.
We have then:
x: weight of the big box
y: weight of the small box.
We now write the system of equations that models the problem:
We apply the graphic solution.
For this, the solution will be the cut point of both lines.
We have then that the solution occurs for the point:
Note: see attached image
Answer:
Weight of each large box: 15.5 Kilograms
Weight of each small box: 13.25 Kilograms
Answer:
0.92
Step-by-step explanation:
always equals 1.0
1.0 - 0.08 = 0.92
Answer:
Step-by-step explanation:
In choices a and b, the bases are positive numbers greater than 1, and so these are growth functions. In c and d, the bases are between 0 and 1, and thus these are decay functions.
In the second problem we have 3ln(x + 1). Rewrite this as ln(x + 1)^3.
We also have 9ln(x - 4). Rewrite this as ln(x - 4)^9.
Because of the + sign connecting ln(x + 1)^3 and ln(x - 4)^9, these two logs combine to form
ln [ (x + 1)^3 ] * (x - 4)^9 (the log of a product).
Now we have:
ln [ (x + 1)^3 ] * (x - 4)^9 - 4ln(x + 7), or:
[ (x + 1)^3 ] * (x - 4)^9
ln ------------------------------------
(x + 7)^9
The answer would be sas or conguernt angle
