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1 year ago

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.The possible outcomes for rolling one six-sided die are shown below. Molly rolled a six-sided die 20 times. The numbers she rol

led are shown below. What is the probability that Molly will roll a one on her next roll? A1/2. B.1/20 C.1/6 D.1/5 https://app136.studyisland.com/pics/121474dicefronts1.pnghttps://app136.studyisland.com/pics/121474dicefront1.pnghttps://app136.studyisland.com/pics/121474dicefront2.png

1 answer:

e-lub [12.9K]1 year ago

3 0

B is the answer 1/20

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Aubree went shopping at her favorite store, Target. She bought 3 shirts and a pair of jeans that cost $28. She spent a total of

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Answer:

88.00 and subtract 28 and the 3 shirts were

Step-by-step explanation:

answer 60

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1 year ago

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Bethany can mow her family’s lawn in 4 hours. Her brother Colin can mow the lawn in 3 hours. Which equation can be used to find

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The answer to this question is C

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1 year ago

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1. tentukan hasil dari 243⅔?

Vesnalui [34]

Penjelasan langkah demi langkah:

1)

$= 243^{\frac{2}{3} }\\= (\sqrt[3]{243})^2\\= 7^2\\= 49$

2) √32 +3√18-2√50

= √16*2 +3√9*2-2√25*2

= 4√2 + 3(3√2)-2(5√2)

= 4√2 + 9√2-10√2

= 13√2-10√2

= 3√2

3) 1000 ⅔×64⅙

$= (\sqrt[3]{1000}) ^2 \times (2^6)^{1/6} \\= 10^2 \times 2\\= 100 \times 2\\= 200$

4) 3/4+√2

$3/4+\sqrt{2} \\= \frac{3+4\sqrt{2} }{4 }$

5) 2√3×√18

= 2√3×√9*2

= 2√3×3√2

= (2*3)(√3*√2)

= 6√6

6) 12/3+√3

= 4+√3

7) √1000—2√40

= 10 -2 (√4*10)

= 10-2(2√10)

= 10 - 4√10

8) 2- ¹+3-¹

$= \frac{1}{2} + \frac{1}{3}\\ = \frac{3+2}{6}\\ = \frac{5}{6}$

9)

$\frac{5}{\sqrt{5} }\\ merasionalisasikan\\= \frac{5}{\sqrt{5} }\times \frac{\sqrt{5} }{\sqrt{5} }\\= \frac{5\sqrt{5} }{\sqrt{25} }\\= \frac{5\sqrt{5} }{5}\\ = \frac{\sqrt{5} }{1}$

Jika pernyataannya opsional, penyebutnya adalah 1

10) 2√3×√18

= 2√3×√9*2

= 2√3×3√2

= (2*3)(√3*√2)

= 6√6

7 0

2 years ago

The price per night for a hotel room is different at three different hotels.

Damm [24]

I think you meant it to be not repeating 3 times so
You do 192/3 is 60*3= 180 leaving you with 12 which is 4*3. So 64 is A
Then it’s 300/5 which is 60
455/7. So first see, if I multiply 7 by 60 is it over or under. If it’s over then B is the least and if it is less then C is the least. So 7 *60 is 420
C Being greater, b costs least per night

4 0

1 year ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago