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3 years ago

7

If the ages in a classroom were 11, 12, 12, 13, 13, 14 - what would the median age be if you added someone that was 10 years old

?

1 answer:

Mandarinka [93]3 years ago

5 0

Answer: 12 years old

Step-by-step explanation:

10,11,12,12,13,13,14

The middle in these numbers is 12

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The answer is B

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100(.25)=27.5
110-27.5=82.5

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3 years ago

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An apartment complex on Ferenginar with 250 units currently has 223 occupants. The current rent for a unit is 892 slips of Gold-

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Answer:

Step-by-step explanation:

Given data

Total units = 250

Current occupants = 223

Rent per unit = 892 slips of Gold-Pressed latinum

Current rent = 892 x 223 =198,916 slips of Gold-Pressed latinum

After increase in the rent, then the rent function becomes

Let us conside 'y' is increased in amount of rent

Then occupants left will be [223 - y]

Rent = [892 + 2y][223 - y] = R[y]

To maximize rent =

$\frac{dR}{dy}=0\\=2(223-y)-(892+2y)=0\\=446-2y-892-2y=0\\=-446-4y=0\\y=\frac{-446}{4}=-111.5$

Since 'y' comes in negative, the owner must decrease his rent to maximixe profit.

Since there are only 250 units available;

$y=-250+223=-27\\\\maximum \,profit =[892+2(-27)][223+27]\\=838 * 250\\=838\,for\,250\,units$

Optimal rent - 838 slips of Gold-Pressed latinum

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4 years ago

The mean grade in Mathematics of Grade 11 students from school A is 83 with standard deviation of 4, while the mean grade in Mat

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Using the context surrounding the word trough in line 9 of "Concrete Mixers," explain what a trough looks like and what it does on a concrete mixer. .

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3 years ago

If (y) varies directly with x, and y=3 when x=0.2, what is the value of (y) when x=1?

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since it varies directly, we form an equation ,
Y =kX where k is a constant
use the points given to find K ,
3 =0.2k , k = 15

So to get final answer,

when x = 1
y = kx, = 1(15)
y= 15

Hence when x=1, < y = 15 >

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3 years ago

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Find the particular solution to y ' = 4sin(x) given the general solution is y = c - 4cos(x) and the initial condition y of pi ov

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The general solutions always have some additive/multiplicative constant, that you must fix in the particular solution.

In order to do so, you need to impose that the particular solution passes through a certain point. In your case, you have

$y(x) = c-4\cos(x)$

and you want

$y\left(\dfrac{\pi}{2}\right) = 2$

Put everything together, and you have

$y\left(\dfrac{\pi}{2}\right) = c-4\cos\left(\dfrac{\pi}{2}\right) = c = 2$

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3 years ago