The inter-quartile selection is 13.
1. Bespeak the data from least to greatest
2. Uncover the mean of the data.
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3. Calculate the typical of both the lower and also upper fifty percent of the data.
4. The inter-quartile selection is the difference in between the upper and lower medians.
Lower Median: 17
Upper Median: 30
Inter-quartile range: 30 - 17 = 13
A sample that 1700 computer system chips revealed that 73% that the chips perform not fail in the first 1000 hrs of their use. The company"s p
It is a left tailed test.
Given : n = 1700
α = 0.05
We desire to test,
the null hypothesis,
the alternate hypothesis,
Test statistics is
Thus z - vital value = 1.64
Since, z calculation ----- z-critical value
Thus we disapprove the null hypothesis.
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So there is adequate evidence in ~ 0.05 level to assistance the company"s case that an ext the 70% perform not fail in the very first 1000 hours of their use.
x = 8
0.4x - 1.2 = 0.15x + 0.8
0.4x - 1.2 + 1.2 = 0.4x
0.15x + 0.8 + 1.2 = 0.15x + 2
0.4x = 0.15x + 2
0.4x - 0.15x = 0.25x
0.15x + 2 - 0.15x = 2
0.25x = 2
0.25x / 0.25 = x
2 / 0.25 = 8
x = 8
Option C (f(x) =
In this question, the very first step is to write the general form of the quadratic equation, i m sorry is f(x) =