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56
99-43
subtract least from greatest
Answer:
n > 96
Therefore, the number of samples should be more than 96 for the width of their confidence interval to be no more than 10mg
Step-by-step explanation:
Given;
Standard deviation r= 25mg
Width of confidence interval w= 10mg
Confidence interval of 95%
Margin of error E = w/2 = 10mg/2 = 5mg
Z at 95% = 1.96
Margin of error E = Z(r/√n)
n = (Z×r/E)^2
n = (1.96 × 25/5)^2
n = (9.8)^2
n = 96.04
n > 96
Therefore, the number of samples should be more than 96 for the width of their confidence interval to be no more than 10mg
Answer:
You could just replace all the given possible values of k in the inequality and see which ones are solutions, but let's solve this in a more interesting way:
First, remember how the absolute value works:
IxI = x if x ≥ 0
IxI = -x if x ≤ 0
Then if we have something like:
IxI < B
We can rewrite this as
-B < x < B
Now let's answer the question, here we have the inequality:
I-k -2I < 18
Then we can rewrite this as:
-18 < (-k - 2) < 18
Now let's isolate k:
first, we can add 2 in the 3 parts of the inequality:
-18 + 2 < -k - 2 + 2 < 18 + 2
-16 < -k < 20
Now we can multiply all sides by -1, remember that this also changes the direction of the signs, then:
-1*-16 > -1*-k > -1*20
16 > k > -20
Then k can be any value between these two limits.
So the correct options (from the given ones) are:
k = -16
k = -8
k = 0