Binomial probability states that the probability of x successes on n repeated trials in an experiment which has two possible outcomes can be obtained by
(nCx).(p^x)⋅((1−p)^(n−x))
Where success on an individual trial is represented by p.
In the given question, obtaining heads in a trial is the success whose probability is 1/2.
Probability of 6 heads with 6 trials = (6C6).((1/2)^6).((1/2)^(6–6))
= 1/(2^6)
= 1/64
2x²-3xy-2y²-2x-11y-12
Write -3xy as a difference
2x²+xy-4xy-2y²-2x-11y-12
write -2x as a difference
2x²+xy-4xy-2y²+4x-6x-11y-12
Write -11y as a difference
2x²+xy-4xy-2y²+4x-6x-8y-3y-12
Factor out 2x from the expression
2x×(x-2y-3)+xy-2y²+4x-8y-3y-12
Factor out y from the expression
2x×(x-2y-3)+y×(x-2y-3)+4x-8y-12
Factor out 4 from the expression
2x×(x-2y-3)+y×(x-2y-3)+4(x-2y-3)
Factor out x-2y-3 from the expression
Answer: (x-2y-3)x(2x+y+4)
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Answer: 198 ft^2
total Area = Area of rectangle + area of trapzoid
= base*height + .5(b1+b2)*height
9*18+.5(9+3)6
162+36
198
Answer:
Here's link to the answer:
bittuly/3a8Nt8
Step-by-step explanation:
Answer:
We accept H₀. We do not have argument to keep the claim that the mean breaking strength has increased
Step-by-step explanation:
Normal Distribution
Population Mean μ₀ = 1850 pounds
Standard Deviation σ = 90 pounds
Type of test
Null Hypothesis H₀ ⇒ μ = μ₀
Alternative Hypothesis Hₐ ⇒ μ > μ₀
A one tail test (right)
n = 21 as n < 30 we use t-student table
degree of fredom 20
t = 2.845
Sample mean μ = 1893
Then, we compute t statistics
t(s) = [ 1893 - 1850 ] / 90/ √n
t(s) = 43 * 4,583 / 90
t(s) = 197,069 / 90
t(s) = 2,190
And we compare t and t(s)
t(s) = 2.190
t = 2.845
Then t(s) < t
We are in the acceptance zone, we accept H₀
