= 0.3133
Given: μ = 560
σ = 90
To find: P (490 < X < 560)
Normal distributions are symmetric, unimodal, and asymptotic. A normal distribution is determined by two parameters the mean and the variance. A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution. It's always easy to solve questions in terms to standard normal
Hence, converting normal distribution to standard normal gives:
P(490 < X < 560) = ≤
= P(0 ≤ Z < 0.888)
= P (z<0.89) - P(z ≤ 0)
Using standard normal table,
= 0.8133 - 0.5
P(490 < X < 560) = 0.3133
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Note that f(1) = 6(1)+2 = 8. Next, evaluate g(x) at x=8: g(f(1)) = g(8) = 2(8) + 4/5, or 16 4/5.
width: x/2, Length:x
18=x(x/2), 18=x^2/2, 36 = x^2, x= -6,6 Use the positive value of 6 for x. the width will be 6/2=3.
Answer:
About 4.8
Step-by-step explanation:
The formula for the area is pi(r^2).
Plugging the area in to the equation, you get 72.382=pi(r^2).
Dividing both sides by pi, you get about 23.04=r^2.
Taking the square root of both sides you get about r=4.8
Answer:
y=x/2-1/4
Step-by-step explanation:
From exercise we have
C=0.
dy/dx+2y=x
Use the formula:
∫xe^(2x)dx=e^(2x)(x/2−1/4).
We know that a linear differential equation is written in the standard form:
y' + a(x)y = f(x)
we get that: a(x)=2 and f(x)=x.
We know that the integrating factor is defined by the formula:
u(x)=e^{∫ a(x) dx}
⇒ u(x)=e^{∫ 2 dx} = e^{2x}
The general solution of the differential equation is in the form:
y=\frac{ ∫ u(x) f(x) dx +C}{u(x)}
⇒ y=\frac{ ∫ e^{2x}· x dx + 0}{e^{2x}}
y=\frac{e^{2x} (x/2-1/4)}{e^{2x}
y=x/2-1/4