Answer:
] y=x-4
-2x+y=18
y - x = -4
y - 2x = 18
equation [2] for the variable y
[2] y = 2x + 18
// Plug this in for variable y in equation [1]
[1] (2x+18) - x = -4
[1] x = -22
// Solve equation [1] for the variable x
[1] x = - 22
// By now we know this much :
y = 2x+18
x = -22
// Use the x value to solve for y
y = 2(-22)+18 = -26
Step-by-step explanation:
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Answer:
And if we solve for a we got
The 95th percentile of the hip breadth of adult men is 16.2 inches.
Step-by-step explanation:
Let X the random variable that represent the hips breadths of a population, and for this case we know the distribution for X is given by:
Where 
and 
For this part we want to find a value a, such that we satisfy this condition:
(a)
(b)
We can find a quantile in the normal standard distribution who accumulates 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64
Using this value we can set up the following equation:
And we have:
And if we solve for a we got
The 95th percentile of the hip breadth of adult men is 16.2 inches.
Answer:
The solution to the inequality is 0.
Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
Answer:
n=-13, n=12
Step-by-step explanation:
