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3 years ago

What is the perimeter of the figure

Mathematics

1 answer:

anastassius [24]3 years ago

6 0

Answer:

Option A.

$P=73.4\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

In this problem we have a triangle with a circle inscribed in it

$A.F=A.B\\C.B=C.D\\E.F=E.D$

step 1

Find the length side B.C

$C.B=A.C-A.B$

substitute the given values

$C.B=23-14=9\ units$

step 2

Find the length side C.D

Remember that

$C.D=C.B$

therefore

$C.D=9\ units$

step 3

Find the length side E.D

we know that

$E.D=E.C-C.D$

substitute the given values

$E.D=22.7-9=13.7\ units$

step 4

Find the length side E.F

Remember that

$E.F=E.D$

therefore

$E.F=13.7\ units$

step 5

Find the length side A.F

Remember that

$A.F=A.B$

therefore

$A.F=14\ units$

step 6

Find the perimeter

$P=23+22.7+13.7+14$

$P=73.4\ units$

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a. {1,2,3,4,5,6}

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Step-by-step explanation:

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As two coins are tossed, the outcomes will be a combination of two outcomes

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d. Rolling a 6 sided die and tossing a coin

As the die is rolled and then a coin is tossed. The die's coin will either combine with heads or tails

So,

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Keywords: Sample Space, Probability

Learn more about sample space at:

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4 years ago

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strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

2. Number of 6-bit strings with weight of 0 is 1

3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

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3 years ago