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vesna_86 [32]
2 years ago
5

Find the area of the polygon. 7 cm 14 cm 14 cm 7 cm 7 cm 7 cm

Mathematics
1 answer:
kvasek [131]2 years ago
7 0

The area of the polygon is 470596 units²

<h3>What is Area of Regular polygon?</h3>

The region occupied by it in a two-dimensional plane. The areas or formulas for areas of different types of polygon depends on their shapes.

Given that: side are 7 cm,14 cm,14 cm,7 cm,7 cm,7 cm

Area of Polygon= 7*14*14*7*7*7

                             = 343*195*7

                             = 470596 units²

The area of polygon is  470596 units²

Learn more about area of polygon here:

brainly.com/question/10761744

#SPJ1

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A Circular pie with a diameter of 10” Is cut into six equal pieces. What is the surface area of two slices of pie?
-Dominant- [34]

To solve this question, let's find the surface area of the entire pie and then divide it by 6 for each slice.

The area of a circle is pi*r^2. We are given the diameter, so divide that by 2 to get the radius.



Now that you have the area, divide by 6. That is the area of each slice. The problem asks for the area of 2 slices though, so multiply that by 2.

4 0
2 years ago
How do u do mean median range
liq [111]
Mean: Add up all of the values then divide by the number of values
Ex.  3+5+2 = 10 divided by three

3 0
4 years ago
Read 2 more answers
Help with both pls ty <br> Pls hurry Tysm!!
maksim [4K]

Answer:

1. 98 cm^2  

2. 471.25

Step-by-step explanation:

5 0
3 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
One-third of the class grew peas, one-third grew carrots, and one-third grew beans. The class consisted of 63 students of which
olga_2 [115]

The number of boys and girls assigned to each team is 6 and 15 respectively.

<h3>Fraction</h3>
  • Number of students in class = 63

  • Number of students who grows peas = 1/3 × 63

= 63/3

= 21 students

  • Number of students who grows carrots = 21
  • Number of students who grows beans = 21

Number of girls = 5/7 × 63

= 315/7

Number of girls = 45 students

Number of boys = 63 - 45 = 18 students

If the three groups have equal composition of boys and girls, then,

Number of boys and girls in each group is 6 and 15 respectively.

Learn more about fraction:

brainly.com/question/11562149

#SPJ1

7 0
3 years ago
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