C, because the graph titles.
Answer:
the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)
Step-by-step explanation:
since the variance S² of the batch follows a normal distribution , then for a sample n of 20 distributions , then the random variable Z:
Z= S²*(n-1)/σ²
follows a χ² ( chi-squared) distribution with (n-1) degrees of freedom
since
S² > 3.10 , σ²= 1.75 , n= 20
thus
Z > 33.65
then from χ² distribution tables:
P(Z > 33.65) = 0.02020
therefore the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)
Answer: see below
Step-by-step explanation:
(i)
200/2 = 100
Find 100 on cumulative frequency and the correspinding time: 14 minutes. So the answer is 14 minutes.
(ii)
First, find the lower quartile —> 200/4 = 50
Upper quartile —> 200/(3/4) = 150
Find the difference: 150 - 50 = 100
So the interquartile range is 100
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Answer:
p = 11.2
Step-by-step explanation:
The computation is shown below:
Data provided in the question
2.6(5.5p – 12.4) = 127.92
Now
Distributive Propertyis
14.3p - 32.24 = 127.92
Addition Property is
14.3p = 127.92 + 32.24
Division Property is
14.3p ÷ 14.3 = 160.16 ÷ 14.3
p = 11.2
We simply find the value of p by applying the distributive property, addition property, and the division property and the same is to be considered
Answer:
these three apply:
ΔCBA ≈ ΔFED
ΔBAC ≈ ΔEDF
ΔABC ≈ ΔDEF
Step-by-step explanation:
