All categories

2 years ago

What rule or property justifies line B.

Mathematics

1 answer:

Goryan [66]2 years ago

5 0

I’m pretty sure it’s number two in my opinion

You might be interested in

Write this number in expanded form twenty three thousand four hundred thirty six

jok3333 [9.3K]

20,000 + 3,000 + 400 + 30 + 6

5 0

2 years ago

Read 2 more answers

PLEASE NEED HELP NOW!!!!!!!❤️

alexira [117]

Answer:

A. 2564 cm^2

Step-by-step explanation:

There are two triangles (Which both are the same) and three different rectangles.

1. Let's solve the area of the triangles first!

Formula: bh(1/2)

10(26)(1/2)

130

2. Now, let's multiply by 2 since there's two of them:

130(2) = 260

3. So, we got 260. Let's find the area of the three rectangles!

Formula: bh

Rectangle A: 28(36) = 1008

Rectangle B: 10(36) = 360

Rectangle C: 26(36) = 936

4. Add the area of the triangles and the rectangles!

260 + 1008 + 360 + 936 = 2564

So the surface area is 2564 cm^2

Hope this helps! <3

4 0

2 years ago

1. There are two differnt maps of Ohio. The scale on the first map is 1 cm to 10 km. The distance from Cleveland to Cincinnati i

xeze [42]

Answer:

1. 0.8 cm

2. 1.6 cm

Step-by-step explanation:

The scale for 2nd map is 1 cm to 50 km, that means "1 cm on map" is "50 km in real life".

We already know distance from Cleveland to Cincinnati is 40 km, which is less than 50, so we know the distance on map would be less than 1 cm.

So we set up ratio and figure out (let x be distance on map from Cleveland to Cincinnati):

$\frac{1}{50}=\frac{x}{40}\\50x=40\\x=\frac{40}{50}\\x=0.8$

Hene, 0.8 centimeters would be the distance in 2nd map

A scale of 1:50 means 1 cm equal 50 cm

So, 0.8m would be

0.8 * 100 = 80 cm

Hence, 80 cm would be represented by 80/50 on the map, that is:

$\frac{80}{50}=1.6$

That is 1.6 centimeters

7 0

2 years ago

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

2 years ago

The measure of ∠FHJ is 60°. The measure of ∠FGH is 28°. What is the measure of ∠HFG?

Hatshy [7]

Given
∠FHJ = 60°
∠FGH = 28°

∠FHG = 180 - ∠FHJ
∠FHG = 180 - 60
∠FHG = 120

∠HFG = 180 - (28 + 120)
∠HFG = 180 - 148
∠HFG = 32

6 0

3 years ago