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3 years ago

Sarah is building a 10-foot-long pen for her pet rabbits. She wants to make gravel bases for two circular food containers

Mathematics

1 answer:

TEA [102]3 years ago

7 0

Answer:

y ≥ 4x

y ≤ 8.75 –1/2πx^2

Step-by-step explanation:

Because the diameters of the gravel bases added together cannot exceed the width of the pen, we get the inequality 2x + 2x ≤ y . Rewriting, we get y ≥ 4x as the first inequality in the system.

Next, write an inequality for cost.

To write the expression for the cost of the fencing, find the perimeter of the rectangle, and multiply the perimeter by the cost per foot of fencing. The pen is a rectangle, so the perimeter is 2(10) + 2(y), or 20 + 2y. Multiply the cost of the fencing material ($4.00 per foot) by the perimeter of the fence to get 4(20 + 2y).

Now, write an expression for the gravel bases for the circular food containers. Because A = r2 and the cost of the gravel is $2.00 per square foot, multiply the cost of the material by the sum of these areas to get 2(x2) + 2(x2).

The total cost must be less than or equal to $150. So, we can say that 4(20 + 2y) + 2(x2) + 2(x2) ≤ 150. After simplifying and solving for y: y ≤ 8.75 – x2.

So, this is the system:

y ≥ 4x

y ≤ 8.75 –1/2πx^2

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Dairy farmers are aware there is often a linear relationship between the age, in years, of a dairy cow and the amount of milk pr

Jobisdone [24]

Answer:

B. A cow of 5 years is predicted to produce 5.5 more gallons per week.

Step-by-step explanation:

Let $M(a) = 40.8-1.1\cdot a$ , where $a$ is the age of the dairy cow, measured in years, and $M(a)$ is the predicted milk production, measured in gallons per week.

Besides, we consider $a_{1}$ and $a_{2}$ , such that $a_{1}\ne a_{2}$ , we define the difference between predicted milk productions ( $\Delta M$ ) below:

$\Delta M = -1.1\cdot (a_{2}-a_{1})$ (1)

If we know that $a_{1} = 5\,yr$ and $a_{2} = 10\,yr$ , then the difference between predicted milk productions is:

$\Delta M = -1.1\cdot (10-5)$

$\Delta M = -5.5\,\frac{gal}{week}$

That is, a cow of 5 years is predicted to produce 5.5 more gallons per week than a cow of 10 years. Hence, the right answer is B.

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3 years ago

Exhibit B: A restaurant has tracked the number of meals served at lunch over the last four weeks. The data shows little in terms

Mekhanik [1.2K]

Answer:

Option E is correct.

The expected number of meals expected to be served on Wednesday in week 5 = 74.2

Step-by-step Explanation:

We will use the data from the four weeks to obtain the fraction of total days that number of meals served at lunch on a Wednesday take and then subsequently check the expected number of meals served at lunch the next Wednesday.

Week

Day 1 2 3 4 | Total

Sunday 40 35 39 43 | 157

Monday 54 55 51 59 | 219

Tuesday 61 60 65 64 | 250

Wednesday 72 77 78 69 | 296

Thursday 89 80 81 79 | 329

Friday 91 90 99 95 | 375

Saturday 80 82 81 83 | 326

Total number of meals served at lunch over the 4 weeks = (157+219+250+296+329+375+326) = 1952

Total number of meals served at lunch on Wednesdays over the 4 weeks = 296

Fraction of total number of meals served at lunch over four weeks that were served on Wednesdays = (296/1952) = 0.1516393443

Total number of meals expected to be served in week 5 = 490

Number of meals expected to be served on Wednesday in week 5 = 0.1516393443 × 490 = 74.3

Checking the options,

74.3 ≈ 74.2

Hence, the expected number of meals expected to be served on Wednesday in week 5 = 74.2

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3 years ago

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<span>Exactly 8*pi - 16
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3 years ago

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3 years ago

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