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Ivan
2 years ago
10

You want to draw a red then a blue marble. do you have a better chance of drawing a red then blue marble with or without replaci

ng the first marble? explain your answer
Mathematics
1 answer:
elena-s [515]2 years ago
7 0

Answer: Without replacing

Step-by-step explanation:

If you don't replace the first red marble, it decreases the number of red marbles in the mix, making the proportion of blue marbles relatively larger. For example, if there were 5 red and 5 blue marbles, there would be a 50% chance of drawing a blue marble. If the red drawn is not replaced, there is a 5/9 chance of drawing a blue marble, or a 55% chance of drawing one.

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The squirrel is 3.16 meters away from the ground, or 316 centimeters.
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Shirley spelled 12 of her 16 spelling words correctly on the test. What percent grade did shirley get on the test
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75%

Step-by-step explanation:

percent = part/whole * 100%

percent = 12/16 * 100%

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Zahra gave out a survey to some students in her school about their favorite color. Students could choose between purple and red.
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2 years ago
What do you call an event where it doesn’t effect the other?<br> A. Dependent<br> B. Independent
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Read 2 more answers
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
Bess [88]

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

Here’s the diagram:

3 0
3 years ago
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