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DochEvi [55]
3 years ago
14

The dataset below shows the number of cars parked in the restaurant parking lot during the lunch hour each day for two weeks:

Mathematics
2 answers:
docker41 [41]3 years ago
5 0

Answer:  C: There is one outlier that indicates an unusually large number of cars were in the parking lot that day.

Step-by-step explanation:

OK so here is our data: 8 7 14 10 13 27 11 10 14 7 12 9 14 9

the simple thing to do it reorganize it from greatest to least

(7,8,10,11,12,14,14,14, 27) as you can see, most of the numbers are 1 or two values apart, but 27 is a whooping 12 more than 14

And since 27 is way more than the rest, we can conclude that the number of cars is unusually large

Katena32 [7]3 years ago
3 0
<span>C. On the 6th day, the cars parked in the lot were 27, which is the highest figure in the entire two weeks period. The other days the numbers varied with slight differences which is normal. However, the difference between the 6th day and the next day which recorded the highest number is very big. A difference of 13 cars is quiet large.</span>
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Step-by-step explanation:

6 0
3 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
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
How many pairs of whole numbers have a sum of 40
Alex
For the writer, <span><span>there are 20
1.       </span>1 + 39 = 40</span> <span><span>
2.       </span>2 + 38 = 40</span>
<span><span>3.       </span>3 + 37 = 40</span>
<span><span>4.       </span>4 + 36 = 40</span> <span><span>
5.       </span>5 + 35 = 40</span> <span><span>
6.       </span>6 + 34 = 40</span>
<span><span>7.       </span>7 + 33 = 40</span>
<span><span>8.       </span>8 + 32 = 40</span> <span><span>
9.       </span>9 + 31 = 40</span>
<span><span>10.   </span>10 + 30 = 40</span> <span><span>
11.   </span>11 + 29 = 40</span> <span><span>
12.   </span>12 + 28 = 40</span>
<span><span>13.   </span>13 + 27 = 40</span> <span><span>
14.   </span>14 + 26 = 40</span> <span><span>
15.   </span>15 + 25 = 40</span> <span><span>
16.   </span>16 + 24 = 40</span> <span><span>
17.   </span>17 + 23 = 40</span> <span><span>
18.   </span>18 + 22 = 40</span> <span><span>
19.   </span>19 + 21 = 40</span> <span><span>
20.   </span>20 + 20 = 40</span>



5 0
3 years ago
A parabola has its vertex at (-6, 5) and focus at (-6.25, 5). Which of the following is an equation for this parabola?
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Answer:

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vertex and focus are horizontally aligned, so parabola is horizontal.

focal length p = distance between focus and vertex = 0.25

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x = a(y-k)² + h,

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a = -1/(4p)

vertex (-6,5)

h = -6

k = 5

p = 0.25

a = -1/(4·0.25) = -1

x = -(y-5)² - 6

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2 years ago
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loris [4]
I think an equation for this could be 5x+6=y
This can be explained as x representing how much the five friends got and y equaling the total. Hope this helps!
8 0
2 years ago
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