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3 years ago

What fraction lie between1/3 and 1/2

Mathematics

2 answers:

PSYCHO15rus [73]3 years ago

5 0

Answer:

1/2.5

Step-by-step explanation:

Alexus [3.1K]3 years ago

4 0

Answer:

2/5

Step-by-step explanation:

1/3 = .3333

1/2 = .5

.4 is in between

.40 = 40/100 = 4/10 = 2/5

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Car traveling at 262.5 miles in 31/2 hours what was the car speed

lidiya [134]

Its 75 miles per hour hope you understand

6 0

3 years ago

Evaluate 2(L+W) for L=10 and W= 2

arsen [322]

Answer:

Step-by-step explanation:

2(10+2)

turns into 2×12

so, that would be 24

5 0

1 year ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago

The probability that a certain machine will produce a defective item is 0.20. if a random sample of 20 items is taken from the o

GenaCL600 [577]

Use binomial distribution, with p=0.20, n=20, x=3
P(X=x)=C(n,x)p^x (1-p)^(n-x)

P(X>=3)
=1-(P(X=0)+P(X=1)+P(X=2))
=1-(C(20,0)0.2^0 (0.8)^(20-0)+C(20,1)0.2^1 (0.8)^(20-1)+C(20,2)0.2^2 (0.8)^(20-2))
=1-(0.0115292+0.057646+0.136909)
=1-0.206085
=0.793915

7 0

3 years ago

X3(2x2 + 3x) a. 2x5 + 3x4 b. 5x9 c. 2x6 + 3x3 d. 5x

Serhud [2]

Answer:

$2x^5+3x^4$