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Lynna [10]
2 years ago
10

A roll of ribbon is used to tie 12 gifts which will be given to teachers on Teacher's

Mathematics
1 answer:
Wewaii [24]2 years ago
3 0

Answer:

22.2m

Step-by-step explanation:

1 roll to 12 gifts

one gifts uses 1,85m of ribbon

1,85m is the length

12 × 1.85 = 22.2m

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Can u guys help pls?
Contact [7]

Answer:

(2 {n}^{2} ) ^{4}  \\ 2 \times 2 \times 2 \times 2 \times  {n}^{2 \times 4}  \\  {2}^{4}  \times  {n}^{8}  \\  = 16 {n}^{8}

<h3>Answer B is correct</h3>

7 0
3 years ago
Hill’s garden has lots of apple trees. 140 apples fall on ground in 35 seconds. How many apples fall in one second?
IRISSAK [1]
4 Apple would fall in 1 second
3 0
3 years ago
In the figure, the ratio of the perimeter of rectangle ABDE to the perimeter of triangle BCD is 2/3 1 3/2 2 . The area of polygo
adelina 88 [10]
It would b 2/3 and the other one 2/4

6 0
3 years ago
(12 divided by 3) 2 – (18 divided by 3 2)) &lt; (4 divided by 2)
gladu [14]

Answer:

(12 \div 3)(2 - (18 \div 32) \\  = 4(2 -  \frac{19}{16} ) \\  = 4( \frac{23}{16} ) \\  =  \frac{23}{4}  \\ since \: (4 \div 2) = 2 \\  \frac{23}{4} is \: not \:  < 2 \: but \: its \:  > 2

4 0
3 years ago
Software to detect fraud in consumer phone cards tracks the number of metropolitan areas where calls originate each day. It is f
pashok25 [27]

Answer:

0.999987

Step-by-step explanation:

Given that

The user is a legitimate one = E₁

The user is a fraudulent one = E₂

The same user originates calls from two metropolitan areas  = A

Use Bay's Theorem to solve the problem

P(E₁) = 0.0131% = 0.000131

P(E₂) = 1 - P(E₁)  = 0.999869

P(A/E₁) = 3%  = 0.03

P(A/E₂) = 30% = 0.3

Given a randomly chosen user originates calls from two or more metropolitan, The probability that the user is fraudulent user is :

P(E_2/A)=\frac{P(E_2)\times P(A/E_2)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}

=\frac{(0.999869)(0.3)}{(0.000131)(0.03)+(0.999869)(0.3)}

\frac{0.2999607}{0.00000393+0.2999607}

\frac{0.2999607}{0.29996463}

= 0.999986898 ≈ 0.999987

6 0
3 years ago
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