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mina [271]
3 years ago
6

A hexagon can be divided into how many triangles by drawing all of the diagonals from one vertex?

Mathematics
2 answers:
ki77a [65]3 years ago
4 0
A. 4 triangles are formed

Readme [11.4K]3 years ago
4 0

Answer:

4 triangles are formed.

Step-by-step explanation:

Given a hexagon that can be divided into triangles by drawing all of the diagonals from one vertex.

we have to find the number of triangles formed.

As shown in attachment if we a diagonals from one vertex then only 3 diagonals are drawn which results into 4 triangles.

Hence, 4 triangles are formed by drawing all of the diagonals from one vertex.

Hence, option A is correct.

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When 2/3 of a number is increased by 20, the sum is then halved, the result obtained is the same as 2/3 of the number, increased
aleksklad [387]

\frac{2/3x+20}{2} =\frac{2}{3} x+3\\\frac{2/3x+20}{2} - 3 = \frac{2}{3}x\\\frac{2/3x+20}{2} - \frac{6}{2} = \frac{2}{3}x\\\frac{2/3x+20-6}{2} = \frac{2}{3}x\\\frac{2/3x+14}{2} = \frac{2}{3}x\\2(\frac{2/3x+14}{2}) = 2(\frac{2}{3}x)\\2/3x+14=\frac{4}{3}x\\14=\frac{4}{3}x-\frac{2}{3}x\\14=\frac{2}{3}x\\\frac{14}{1}/\frac{2}{3}=x\\\frac{14}{1}*\frac{3}{2}=x\\7*3=x\\x=21

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4 0
3 years ago
Lots of 40 components each are deemed unacceptable if they contain 3 or more defectives. The procedure for sampling a lot is to
MrRissso [65]

Answer:

The probability of founding exactly one defective item in the sample is P=0.275.

The mean and variance of defective components in the sample are:

\mu=0.375\\\\\sigma^2=0.347

Step-by-step explanation:

In the case we have a lot with 3 defectives components, the proportion of defectives is:

p=3/40=0.075

a) The number of defectives components in the 5-components sample will follow a binomial distribution B(5,0.075).

The probability of having one defective in the sample is:

P(k=5)=\binom{5}{1}p^1(1-p)^4=5*0.075*0.925^4=0.275

b) The mean and variance of defective components in the sample is:

\mu=np=5*0.075=0.375\\\\\sigma^2=npq=5*0.075*0.925=0.347

The Chebyschev's inequality established:

P(|X-\mu|\geq k\sigma)\leq \frac{1}{k^2}

6 0
3 years ago
Expressions that simplify to the same expression are called ____ expressions.
larisa86 [58]
They're called equivalent expressions
5 0
3 years ago
Read 2 more answers
HELP ASAP 50 POINT FOR CORRECT ANSWER
satela [25.4K]
Im not sure but I think it is like 65

5 0
3 years ago
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lidiya [134]

Step-by-step explanation:

if14a^2b/7a^2b answer is 2

and if the sign is addition then the answer is21a^2b if subtraction answer is7a^2b

4 0
3 years ago
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