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anastassius [24]
3 years ago
15

Tom has 12 cards with the numbers 1 to 12 on them.

Mathematics
1 answer:
jarptica [38.1K]3 years ago
5 0

Answer:

pls mark me

Step-by-step explanation:

As there are no 12 cards in a standard pack the probability is zero.

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12 is what percent of 44
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Answer:

27.27 %

Step-by-step explanation:

% = part/whole * 100%

(12/44)*100 ≈ 27. 27%

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3 years ago
You are designing a rectangular poster to contain 7575 in2 of printing with a 33​-in margin at the top and bottom and a 11​-in m
erma4kov [3.2K]

Answer:

The dimensions of the rectangular poster is 15 in by 5 in.

Step-by-step explanation:

Given that, the area of the rectangular poster is 75 in².

Let the length of the rectangular poster be x and the width of the rectangular poster be y.

The area of the poster = xy in².

\therefore xy=75

\Rightarrow y=\frac{75}{x}....(1)

1 in margin at each sides and 3 in margin at top and bottom.

Then the length of printing space is= (x-2.3) in

                                                           =(x-6) in

The width of printing space is = (y-2.1) in

                                                  =(y-2) in

The area of the printing space is A =(x-6)(y-2) in²

∴ A =(x-6)(y-2)

Putting the value of y

\Rightarrow A =(x-6)(\frac{75}{x}-2)

\Rightarrow A = 87-\frac{450}{x}-2x

Differentiating with respect to x

A '= \frac{450}{x^2}-2

Again differentiating with respect to x

A''=-\frac{900}{x^3}

To find the minimum area of printing space, we set A' = 0

\therefore \frac{450}{x^2}-2=0

\Rightarrow 450 =2x^2

\Rightarrow x^2=225

\Rightarrow x=\pm 15

Now putting x=±15 in A''

A''|_{x=15}=-\frac{900}{15^3}

A''|_{x=-15}=-\frac{900}{(-15)^3}=\frac{900}{(15)^3}>0

Since at x=15 , A"<0 Therefore at x=15 , the area will be minimize.

From (1) we get

y=\frac{75}{x}

Putting the value of x

y=\frac{75}{15}

   =5 in

The dimensions of the rectangular poster is 15 in by 5 in.

4 0
3 years ago
What is the greatest number of right angles a triangle can contain?
Firdavs [7]

Answer:

C.0

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