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2 years ago

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Leila bought 3 bananas, which weighed a total of Three-fourths of a pound. If each banana weighed the same amount, what is the w

eight of each banana? StartFraction 1 Over 12 EndFraction of a pound StartFraction 1 Over 9 EndFraction of a pound One-fourth of a pound One-third of a pound

2 answers:

nlexa [21]2 years ago

6 0

Answer:

the guy above me is right my frog moon told me

Step-by-step explanation:

Anna [14]2 years ago

4 0

Answer:

$Each\ Weight = \frac{1}{4}$ lb

Step-by-step explanation:

Given

$Weights = \frac{3}{4}lb$

$Bananas = 3$

Required

Determine the weight of each

To do this, we simply divide the total weight by the number of bananas.

$Each\ Weight = \frac{3}{4}/3$

Convert to multiplication

$Each\ Weight = \frac{3}{4} * \frac{1}{3}$

$Each\ Weight = \frac{1}{4}$

<em>Hence, each banana weighs 1/4 of a pound</em>

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90 men and 60 women were asked if they had watched the latest 'Expendables' movie.

Stolb23 [73]

Answer:

63 men said no.

Step-by-step explanation

3/5 * 2/2 = 6/10, 6/10 - 3/10 (women) = 3/10 meaning 7/10 men said no. 90*1.7 = 153, 153-90=63

5 0

2 years ago

Read 2 more answers

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

On this graph, 4:00 p.m. occurs at 16 hours after midnight, and 6:00 p.m. occurs at 18 hours after midnight. Which statements ar

KatRina [158]

Answer:

TRUE OPTIONS ARE:

<em>"The temperature increased until 4:00 p.m. "</em>

<em>"The temperature decreased after 6:00 p.m. "</em>

<em>"The temperature increased more quickly between 12:00 p.m. and 4:00 p.m. than before 12:00 p.m."</em>

Step-by-step explanation:

<em>graph attached and complete question below:</em>

<em>Which statements are true about the temperatures Luis recorded on the graph? Check all that apply. </em>

<em>The temperature increased until 4:00 p.m. </em>
<em>The temperature was not recorded between 4:00 p.m. and 6:00 p.m. </em>
<em>The temperature decreased after 6:00 p.m. </em>
<em>The temperature increased and then decreased before holding constant. </em>
<em>The temperature increased more quickly between 12:00 p.m. and 4:00 p.m. than before 12:00 p.m.</em>

<em />

Until 1600 hours, the graph increases, so we can say temperature increased until 4.00 pm (FIRST OPTION TRUE).

From 1600 to 1800 hours (4 - 6pm), the temperature stayed same (horizontal line). This doesn't mean the temperature wasn't recorded. (2nd OPTION FALSE).

After 1800 hours (6pm), the line goes downward, so temperature decreased after 6 pm. (3rd OPTION TRUE).

If you look at the temperature graph, we can see temperature increased, then increased more, then constant, then decreased. Thus the 4th option isnt true. (4th OPTION FALSE).

Before 12, the increase isn't as sharp as after 12. After 12 temperature increase has more slope, so this increase is more. (5th OPTION TRUE).

7 0

2 years ago

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Dalia flies an ultralight plane with a tailwind to a nearby town in 1/3 of an hour. On the return trip, she travels the same dis

pashok25 [27]

Dalia had an average airspeed of ⇒ 42 miles per hour.
The average wind speed was ⇒ 12 miles per hour.

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2 years ago

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Jesse made 312 mini energy bars she put 24 bars in each bag she plans to sell each bag for $6

lukranit [14]

Answer:

There will be 13 such bags and Jesse will earn $ 78 by selling the energy bars.

Step-by-step explanation:

Jesse made 312 mini energy bars. She put 24 bars in each bag. So, there will be, $\frac {312}{24}$ = 13 such bags.

She plans to sell each bag for $6, so she will earn

$ $(13 \times 6)$

= $ 78 by selling the energy bars.

8 0

2 years ago