Answer:
1/2.
Step-by-step explanation:
10m = 5n
m = 5n/10
m = n/2
m/n = n/2 * 1/n
m/n = 1/2.
C. It’s goes over two and up three. So therefore L would be (3,2)
Eleven minus four equal seven
11-4=7
The probability that the mean clock life would differ from the population mean by greater than 12.5 years is 98.30%.
Given mean of 14 years, variance of 25 and sample size is 50.
We have to calculate the probability that the mean clock life would differ from the population mean by greater than 1.5 years.
μ=14,
σ=
=5
n=50
s orσ =5/
=0.7071.
This is 1 subtracted by the p value of z when X=12.5.
So,
z=X-μ/σ
=12.5-14/0.7071
=-2.12
P value=0.0170
1-0.0170=0.9830
=98.30%
Hence the probability that the mean clock life would differ from the population mean by greater than 1.5 years is 98.30%.
Learn more about probability at brainly.com/question/24756209
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There is a mistake in question and correct question is as under:
What is the probability that the mean clock life would differ from the population mean by greater than 12.5 years?
Answer:
All non zero digits count in significant digits.
(a) is correct option.
Step-by-step explanation:
Given that,
1.4 has 2 significant digits.
We know that,
Significant digits :
All non zero digits count in significant digits.
All zero which is present in between two significant digits it is also significant.
Leading zero is not count as significant.
Trailing zero is count as significant.
We need to find 1.4 has 2 significant digits as a result of which rule
Using given rules
1.4 has 2 significant digits it is follows the rule of all non zero digits count in significant digits.
Hence, All non zero digits count in significant digits.
(a) is correct option.