I am 64 inches tall to the nearest inch,<br> What is the error interval for my height?

Helpppp me please!! I dont understand.

Dafna1 [17]

Answer:

X = 11

X = 7

Step-by-step explanation:

Part 1:

The measure of an inscribed angle is half the measure of an intercepted arc. So 1/2 mCE = m∠CDE

158 ÷ 2 = 79

You need to find x so you're going to set your equation of (8x - 9) = 79

Then you're going to add 9 to both sides to get 8x = 88

Divide both sides by 8 to get x = 11

Part 2:

We know that WY is 180 because it's a semicircle. Using this and the knowledge that ∠WXY is half the measure of the intercepted arc (180), we know that our equation should be (13x - 1) = 90

add 1 to both sides and our equation is (13x = 91)

Divide both sides by 13 and...

x = 7

8 0

3 years ago

Read 2 more answers

Write a word problem that can be solved by ordering 3 decimals to thousand include a solution

IrinaK [193]

Answer:

I have no clue but this was two years ago so hopefully you know the answer by now

Step-by-step explanation:

yes

6 0

3 years ago

Kai took five tests this semester. His scores are listed below.

Travka [436]

Answer:

3.2 is the answer. Your welcome

3 0

3 years ago

HELP ME OUT HERE! PLS TELL ME HOW U GOT THE ANSWER TOO!

IgorC [24]

Answer: Choice C

$\begin{array}{|c|c|} \cline{1-2}x & y\\\cline{1-2}0 & 1\\\cline{1-2}2 & 2\\\cline{1-2}4 & 3\\\cline{1-2}6 & 4\\\cline{1-2}\end{array}$

=========================================================

Explanation:

There are four marked points on the line.

Each point is of the form (x,y)

The first or left most point is (0,1)
The second point is (2,2)
The third is (4,3)
The fourth is (6,4)

Each of these points is then listed in the table format as shown above.

There are infinitely many other points on the line; however, we only select a few of them to make the table (or else we'd be here all day).

Extra side notes:

The slope of this line is m = 1/2 = 0.5
The y intercept is 1 located at (0,1)
The equation of this line is y = 0.5x+1

8 0

2 years ago

Read 2 more answers

) 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

I am 64 inches tall to the nearest inch, What is the error interval for my height?