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Marianna [84]
3 years ago
10

The weights for newborn babies is approximately normally distributed with a mean of 5.4 pounds and a standard deviation of 1.8 p

ounds. Consider a group of 1500 newborn babies: 1. How many would you expect to weigh between 3 and 6 pounds
Mathematics
1 answer:
Mariana [72]3 years ago
6 0

Answer:

You would expect 807 babies  to weigh between 3 and 6 pounds.

Step-by-step explanation:

We are given that

Mean,\mu=5.4pounds

Standard deviation,\sigma=1.8pounds

n=1500

We have to find how  many would you expect to weigh between 3 and 6 pounds.

The weights for newborn babies is approximately normally distributed.

Now,

P(3

=P(-1.33

P(3

P(3

P(3

Number of newborn  babies expect to weigh between 3 and 6 pounds

=1500\times 0.538=807

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SOLUTION:

tan(A) = \frac{opposite}{adjacent}

Opposite = 60

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tan(A) = \frac{60}{11}

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tan(A) = \frac{2\sqrt{15}}{11}

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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
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Answer: -1.0018*10^42

Step-by-step explanation:

The Euler Method in numerical analysis is used to approximate the solution to an initial value problem using the tangent line to the solution curve through the (x0,y0) to obtain such approximations.

The euler method equation is:

Yn+1 = Yn + h*f(Xn,Yn)

Where n = number of steps, h = Xn+1 - Xn, f(Xn,Yn) is the slope of the curve at (Xn,Yn).

Variables given in the equation

Y(X=0) = 9,

X0 = 0.

h = step = 0.2

f(X,Y) = X^2*Y - 12*Y^2

B

For the first step, n = 0 in the euler equation. Therefore we have:

Y1 = Y0 + h*f(X0,Y0)

Substituting the Y0 = 9 and X0 = 0 into the function f(X,Y) = X^2*Y - 12*Y^2 = 0^2*9 - 12*9^2 = -972.

Therefore Y1 = 9 + 0.2*(-972) = -185.4.

For n = 1

Repeating the same step with X1 = X0 + h = 0 + 0.2 = 0.2,

Y1 = -185.4,

h = 0.2.

Substitute the variables into the equation Y2 = Y1 + h*f(X1,Y1) and f(X1,Y1) = X1^2*Y1 - 12*Y1^2

Y2 = -82682.4672.

Continue the iterations following the steps above till the result is reached.

The summary of the iteration.

When n = 0, X = 0, Y = -185.4

When n = 1 , X = 0.2, Y = -82682.4672

When n = 2 , X = 0.4, Y = -1.6407*10^10

When n = 3, X = 0.6, Y = -6.4609*10^20

When n = 4, X = 0.8, Y = -1.0018*10^42

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2 years ago
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