X-y=3=>x=3+y
X+2y=-6
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3+y+2y=-6
3+3y=-6
3y=-6-3
3y=-9
y=-3
X+2y=-6
X+2(-3)=-6
X-6=-6
X=0
C: none of these are solutions to the given equation.
• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.
• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.
The actual solution is easy to find, since this equation is separable.
<em>y'</em> - <em>y</em> = 0
d<em>y</em>/d<em>x</em> = <em>y</em>
d<em>y</em>/<em>y</em> = d<em>x</em>
∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>
ln|<em>y</em>| = <em>x</em> + <em>C</em>
<em>y</em> = exp(<em>x</em> + <em>C </em>)
<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>
Answer: the probability that a randomly selected Canadian baby is a large baby is 0.19
Step-by-step explanation:
Since the birth weights of babies born in Canada is assumed to be normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = birth weights of babies
µ = mean weight
σ = standard deviation
From the information given,
µ = 3500 grams
σ = 560 grams
We want to find the probability or that a randomly selected Canadian baby is a large baby(weighs more than 4000 grams). It is expressed as
P(x > 4000) = 1 - P(x ≤ 4000)
For x = 4000,
z = (4000 - 3500)/560 = 0.89
Looking at the normal distribution table, the probability corresponding to the z score is 0.81
P(x > 4000) = 1 - 0.81 = 0.19
Answer:
Step-by-step explanation:
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