A consumer affairs investigator records the repair cost for 4 randomly selected washers. A sample mean of $64.26 and standard de
1 answer:
Answer:
Critical value: z = 1.28
The 80% confidence interval for the mean repair cost for the washers is between $46.487 and $82.033.
Step-by-step explanation:
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 1-0.1 = 0.9, so z = 1.28
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
So
The lower end of the interval is the mean subtracted by M. So 64.26 - 17.773 = $46.487.
The upper end of the interval is M added to the mean. So 64.26 + 17.773 = $82.033.
The 80% confidence interval for the mean repair cost for the washers is between $46.487 and $82.033.
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Answer:
Step 1 is incorrect.
Step-by-step explanation:
The given expression is :
In step 1, Deepak rearranged the terms in order to simplify like terms together that is followed in step 2.
o, rearranging of the terms of an expression is commutative property and not distributive property. The distributive property is generally applicable to multiplication and is given as:
The commutative property is given as:
x+y = y +a
Therefore, adding terms by changing their order is commutative property of addition and not distributive property. Step 1 instruction is wrong.
Answer:
(t, t -1, t)
Step-by-step explanation:
You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called. I think it refers to infinite many solutions). Here's how it works:
Set up your matrix:
You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left. Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0. Multiplying in a -2 to the top row gives you:
Then add, keeping the first row the same and changing the second to reflect the addition:
The second equation is this now:
3y - 3z = -3. Solving for y gives you y = z - 1. Let's let z = t (some random real number that will make the system true. Any number will work. I'll show you at the end. Just bear with me...)
lf z = t, and if y = z - 1, then y = t - 1. So far we have that y = t - 1 and z = t. Now we solve for x:
From the first equation in the original system,
x - 2y + z = 2. Subbing in t - 1 for y and t for z:
x - 2(t - 1) + t = 2. Simplify to get
x - 2t + 2 + t = 2 and x - t = 0, and x = t. So the solution set is (t, t - 1, t). Picking a random value for t of, let's say 2, sub that in and make sure it works. If:
x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2. Check it: 2 - 4 + 4 = 2 and 2 = 2. You could pick any value for t and it will work.