Answer:
a
The estimate is 
b
Method B this is because the faulty breaks are less
Step-by-step explanation:
The number of microchips broken in method A is 
The number of faulty breaks of method A is 
The number of microchips broken in method B is 
The number of faulty breaks of method A is 
The proportion of the faulty breaks to the total breaks in method A is


The proportion of the faulty to the total breaks in method B is

For this estimation the standard error is

substituting values


The z-values of confidence coefficient of 0.95 from the z-table is

The difference between proportions of improperly broken microchips for the two breaking methods is mathematically represented as
![K = [p_1 - p_2 ] \pm z_{0.95} * SE](https://tex.z-dn.net/?f=K%20%3D%20%5Bp_1%20-%20p_2%20%5D%20%5Cpm%20z_%7B0.95%7D%20%2A%20SE)
substituting values
![K = [0.08 - 0.07 ] \pm 1.96 *0.0186](https://tex.z-dn.net/?f=K%20%3D%20%5B0.08%20-%200.07%20%5D%20%5Cpm%201.96%20%2A0.0186)

The interval of the difference between proportions of improperly broken microchips for the two breaking methods is

Answer:

Step-by-step explanation:
Given equation of ellipsoids,

The vector normal to the given equation of ellipsoid will be given by





Hence, the unit normal vector can be given by,



Hence, the unit vector normal to each point of the given ellipsoid surface is

Answer:
x=82/81
Step-by-step explanation:
Answer: the associative property of addition
Step-by-step explanation: when adding three or more terms, you can combine the middle terms with either the first term or the last term and as you add the results, the sum will be the same either way.
There will be 50 gift bags.
The first set of 25 bags
----------------------------------
Because there are 25 thumb drives, the first 25 bags will each have 1 thumb drive.
Also, each bag will have 5 times as many key chains as a thumb drive. Therefore each of the first 25 bags will have
1 thumb drive, 5 key chains.
After the first 25 bags, we have used
25 thumb drives, 5*25 = 125 key chains.
The second set of 25 bags
----------------------------------------
We have a total of 200 key chains, so we have 200-125 = 75 key chains left.
Distribute them equally among the remaining 25 bags, so each bag has
75/25 = 3 key chains.
Answer:
25 bags, each with 1 thumb drive and 5 key chains.
Another 25 bags, each with 3 key chains.