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3 years ago

9

A store sells apples in 5-pound bags for $2.25. If the price per pound is the same, how much money would it cost for a 20-pound

bag of apples?

1 answer:

dedylja [7]3 years ago

7 0

I’m pretty sure the answer would be $9.00.
EXPLANATION:
2.25/5= 0.45
0.45x20= 9
ANSWER: $9.00

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Solve the system of equations by graphing, elimination or

vladimir1956 [14]

Answer:

Solve the system of equations by graphing, elimination or

subod

Step-by-step explanation:

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3 years ago

AYUDAAA porfavorrrr primera urna 6 bolas verdes y 3 rojas segunda urna 3 verdes, 3 blancas y 3 rojas tercera urna 6 verdes, 1 bl

marin [14]

We have the following information:

first urn: 6 green balls and 3 red ones

total: 6 + 3 = 9

second urn: 3 green, 3 white and 3 red

total: 3 + 3 + 3 = 9

third urn: 6 green, 1 white and 2 red

total: 6 + 1 + 2 = 9

a) A green ball is more likely to be obtained, since there are more green balls than red balls, which makes the probability higher.

b) probability of drawing a green, red and white ball.

first urn:

green = 6/9 = 66.66%

red = 3/9 = 33.33%

white = 0/9 = 0%

second urn:

green = 3/9 = 33.33%

red = 3/9 = 33.33%

white = 3/9 = 33.33%

third urn:

green = 6/9 = 66.66%

red = 2/9 = 22.22%

white = 1/9 = 11.11%

c) it would be chosen where the probability of drawing green would be the highest, which means that it would be possible both in the first and in the third ballot box, the probability is equal 66.66%

d) without a green ball, the third ballot box would look like this:

5 green balls, 2 red balls and 1 white ball, with a total of 8.

The probability of drawing would be:

green = 5/8 = 62.5%

red = 2/8 = 25%

white = 1/8 = 12.5%

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3 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

VladimirAG [237]

First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:

$y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}$

And the derivative of x:

$x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}$

Now, we can calculate the area of the surface:

$A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)$

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

$(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\$

$=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\$

$=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}$

Calculate indefinite integral:

$\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}$

And the area:

$A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}$

Answer D.

6 0

4 years ago

Read 2 more answers

Line segment AB is drawn in the coordinate plane. Carlos graphed the set of points that are the same distance from both point A

Illusion [34]

Answer:

Carlos graphed the perpendicular bisector of AB.

Step-by-step explanation:

Given a line segment AB drawn in a coordinate plane. The set of points that are the same distance from points A and B divides the line segment AB into two equal parts.

This line is referred to as the perpendicular bisector of AB.

Therefore, Carlos graphed the perpendicular bisector of AB.

6 0

3 years ago

The radius of circle L is 16 cm.

timurjin [86]

The anwser is for the question tht u are looking for is no other then the first anwser A!!!

6 0

3 years ago