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3 years ago

Um simpafly if you can plz :]

Mathematics

2 answers:

user100 [1]3 years ago

7 0

Answer:

um simpafly if you can plz :]

answer is = 15/56

Karolina [17]3 years ago

5 0

Answer:

15/56

Step-by-step explanation:

You just multiply across. It is in the simplest form. Hope this helps, let me know if correct, and have a great day!!

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Please help! Is it 0.0364?

NNADVOKAT [17]

Answer:

0.0636

Step-by-step explanation:

The probability the first die has twelve sides is 4/11.

The probability the second die has twelve sides is 3/10.

The probability the third die does not have twelve sides is 7/9.

The probability the fourth die does not have twelve sides is 6/8.

The probability of all four events is:

4/11 × 3/10 × 7/9 × 6/8 = 0.0636

7 0

4 years ago

Question 4 of 10

stiks02 [169]

Answer:

28° A

Step-by-step explanation:

we have isosceles triangle, which's 2 sides and 2 angles are equal.

we have two equal sides, so another side is base.

base angles are equal, so if E=28 thenG=28 too

3 0

3 years ago

What word describes the relationship between oppsite angles of a parallelogram?

meriva

Congruent........................

5 0

4 years ago

Read 2 more answers

Solve:m+ 2/3=1/2 A. 1 1/6 B. 1/6 C.-1 D. - 1/6

natita [175]

Answer:

D. - 1/6

Step-by-step explanation:

M + 2/3 = 1/2

M = 1/2 - 2/3

$\frac{1}{2} - \frac{2}{3} = \frac{3}{6} - \frac{4}{6} = -\frac{1}{6}$

6 0

3 years ago

Read 2 more answers

The top and bottom margins of a poster are each $3$ cm and the side margins are each $2$ cm. If the area of printed material on

babymother [125]

Answer:

Therefore the dimension of the poster is 12 cm by 8 cm.

Step-by-step explanation:

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

Given that the area of the poster is 96 cm².

∴xy =96

$\Rightarrow y= \frac{96}{x}$

The sides margins each are 2 cm and the top and bottom margins of the poster are each 3 cm.

The length of printing space is =(x- 2.3) cm

= (x-6) cm

The width of the printing space is =(y-2.2) cm

=( y-4 )cm

The area of the printing space is A=(x-6)(y-4) cm²

∴A=(x-6)(y-4)

$\Rightarrow A=(x-6)(\frac{96}{x}-4)$ [ Putting $y= \frac{96}{x}$ ]

$\Rightarrow A=120-\frac{576}{x}-4x$

Differentiating with respect to x

$A'= \frac{576}{x^2}-4$

Again differentiating with respect to x

$A''=-\frac{1152}{x^3}$

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

$\therefore \frac{576}{x^2}-4=0$

$\Rightarrow \frac{576}{x^2}=4$

$\Rightarrow x^2=\frac{576}{4}$

$\Rightarrow x^2 =144$

$\Rightarrow x=\pm 12$

Dimension can't be negative.

Therefore x=12

If x=12, the value of A''>0,then at x=12, the area of the poster will be minimum.

If x=12, the value of A''<0,then at x=12, the area of the poster will be minimum.

$\therefore A''|_{x=12}=-\frac{1152}{12^3}$

Therefore at x= 12 cm the area of the poster will be maximum.

The width of the poster is $y=\frac{96}{12}$ = 8 cm

Therefore the dimension of the poster is 12 cm by 8 cm.

3 0

3 years ago