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3 years ago

In dividing an irregular shape into polygons it is permissible for the polygons to overlap

1 answer:

6 0

NO it is not possible when dividing an irregular shape to polygons to overlap. Every polygon would be broken up in piece.

You might be interested in

What equation represents the graphed function

Marina86 [1]

Answer: $y=\dfrac{-1}{3}x+3$

Step-by-step explanation:

The equation of ab line that passes through two points (a,b) and (c,d) is given by :-

$(y-b)=\dfrac{d-c}{b-a}(x-a)$

From the given picture , the line is passing through points (0,3) and (3,2).

Then, the equation of the line will be :-

$(y-3)=\dfrac{2-3}{3-0}(x-0)$

$\Rightarrow y-3=\dfrac{-1}{3}x$

$\Rightarrow y=\dfrac{-1}{3}x+3$

Hence, the required equation of the line that represents the graphed function : $y=\dfrac{-1}{3}x+3$

7 0

3 years ago

Read 2 more answers

Y = 4x - 6

Studentka2010 [4]

Answer:

B) 2x - 12x - 18 = 8

Step-by-step explanation:

2x - 3y = 8

If y = 4x - 6, then substitute and solve:

2x - 3(4x - 6) = 8

2x - 12x - 18 = 8

10x - 18 = 8

10x = 8 + 18

10x = 26

x = 26/10

x = 2.6

y = 4(2.6) - 6

y = 10.4 - 6

y = 4.4

8 0

3 years ago

Donna is running a track it takes her 10 minutes to run 6 laps if she keeps running at the same speed how long will it take her

kaheart [24]

To solve this problem we can make a proportion.
6 5
-- = ---
10 x

to solve this proportion we can cross multiply and divide.
5 times 10 is 50. divided by 6 is 8.3
it will take her 8.3 minutes to run 5 laps

7 0

3 years ago

Read 2 more answers

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

2 years ago

the length of a side of a square is 10x+4 units. If the perimeter is 96 units. What is the area of the square?

dmitriy555 [2]

Answer:

576 Units

Step-by-step explanation:

The way your teacher wants you to do it:

1. We know that multiplying a length by four gives us the perimeter of a square.

1a. Thus, we multiply the given length by four. 4(10x+4)=40x+16

1b. We know the perimeter is 96. So: 40x+16=96

2. Solving for x, we get x=2

3. Plugging our solved x into the given length (10x+4), we see that the length of a side of the square is 10*2+4=24.

4. We know that the area of a square can be found by squaring the side of a square.

4a. 24*24=576.

5. The area is 576 units.

--------------------------------------------------------------

Easier Solution:

1. We know that the perimeter is length*4.

2. We set L to length.

2a. 4L=96

2b. L=24

3. We know that the length square is area.

4. 24*24=576.

4 0

3 years ago