Answer:
For this case we want to determine if pennies are really fair when flipped, meaning equally likely to land head up or tails, so then the correct system of hypothesis are:
Null hypothesis: 
Alternative hypothesis: 
Step-by-step explanation:
Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Solution to the problem
For this case we want to determine if pennies are really fair when flipped, meaning equally likely to land head up or tails, so then the correct system of hypothesis are:
Null hypothesis:
Alternative hypothesis:
(9,40,41) is a Pythagorean Triple, farther down the list than teachers usually venture.
Answer: D. 41 cm
There's a subset of Pythagorean Triples where the long leg is one less than the hypotenuse,
a^2+b^2 = (b+1)^2
a^2 + b^2 = b^2 + 2b +1
a^2=2b+1
So we get one for every odd number, since the square of an odd number is odd and the square of an even number is even.
b = (a^2 - 1)/2
a=3, b=(3^2-1)/2=4, c=b+1=5
a=5, b=(5^2-1)/2 =12, c = 13
a=7, b=24, c=25
a=9, b=40, c=41
a=11, b=60, c=61
a=13, b=84, c=85
It's good to be able to recognize Pythagorean Triples when we see them.
Otherwise we'd have to work the calculator:
√(9² + 40²) = √1681 = 41
There would be 12 litres in the tub after 4 minutes
Answer:
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 1.1156 occur / millimeters
Step-by-step explanation:
<u>Step 1</u>:-
Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.
by Poisson random variable given that λ = 1.5 flaws/millimeter
Poisson distribution 
<u>Step 2:</u>-
The probability that no flaws occur in a certain portion of wire
On simplification we get
P(x=0) = 0.223 flaws occur / millimeters
<u>Conclusion</u>:-
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters
