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3 years ago

9

A jar contains 10 red marbles and 20 blue marbles. if one red marble is selected and then placed back into the jar, what is the

probability of the next marble being a blue marble?

1 answer:

Serggg [28]3 years ago

6 0

You have a 30% chance of getting a blue marble. <span />

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Given: ∠2 ≅ ∠4 Which lines must be parallel?

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Answer:

B) p and q

Step-by-step explanation:

If ∠2 ≅ ∠4 then line p and line q must be parallel. ∠2 and ∠4 are corresponding angles, and corresponding angles are congruent.

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2x(X-5)+6=24 X=numar necunoscut

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Answer:

2x² + 6 = 34x

Step-by-step explanation:

2x² - 10x + 6 = 24x

Collect like terms

2x² + 6 = 24x + 10x

2x² + 6 = 34x

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4 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

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Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let <em>x</em> be the width and <em>y </em>the length of the rectangular field.

Let <em>C </em>the total cost of the rectangular field.

The side made of heavy duty material of length of <em>x </em>costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are <em>x, y, y</em>. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$

The least cost of fencing for the rancher is $1200

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