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lilavasa [31]
2 years ago
10

Find the area of the shaded section only in the figure below.

Mathematics
2 answers:
Evgen [1.6K]2 years ago
5 0
Aha.. you forgot to attach a photo! its ok i have done that as well!
Vesna [10]2 years ago
5 0
If you show me i can try and help
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AB = x + 8
vivado [14]

Answer: The length of BC is 7

Step-by-step explanation: Assuming the lengths of the opposite sides of the quadrilateral are congruent, then

AB=DC and

AD=BC

Inputting the values of AB, DC and AD as given in the question:

x + 8 = 3x ...(1)

x + 3=? ...(2)

We have to solve for the value of x to get the actual lengths and thus ascertain BD.

From equation (1):

8 = 3x - x

8 = 2x

8/2 = x

Therefore, x = 4.

If x = 4 then equation(2) would be

4 + 3= 7.

Hence, the actual lengths of the quadrilateral are:

AB = 4 + 8. DC = 3(4)

=12. =12.

AD = 4 + 3. AD = BC

= 7. Therefore, BC = 7.

Hence, it is confirmed that quadrilateral ABCD is a parallelogram since both the opposite sides are proven to be congruent.

3 0
3 years ago
HELLPPP PLSSS<br><br>100÷10+3-100+ (6+ 19)+ 3 × -5 -19+100​
Ksivusya [100]

Answer:

4

Step-by-step explanation:

I used a Calculater

3 0
2 years ago
Helppppppppppp please
LenaWriter [7]

Answer:

x = 60°

Step-by-step explanation:

AOB    Center: O      AB: diameter

arc ADB = 180°

arc DB = 180 - arc AD = 80°

x = 1/2 x ( arc DB + arc AC) = 1/2 x (80 + 40) = 60°

3 0
3 years ago
If you are a dog lover, having your dog with you may reduce your stress level. Does having a friend with you reduce stress? To e
Pepsi [2]

The outlier of a dataset is a data element that is relatively far from the remaining data elements

  • <em>99 is an outlier of pet group</em>
  • <em>See attachment for the parallel box plots</em>

<u>(a) Prove that 99 is an outlier for Pet</u>

We have:

<em>Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99</em>

n = 15

The quartiles positions are:

Q_1 = \frac{n + 1}{4}

Q_1 = \frac{15 + 1}{4}

Q_1 = \frac{16}{4}

Q_1 = 4th

Q_3 = Q_1 \times 3

Q_3 = 4th \times 3

Q_3 = 12th

So, we have:

Q_1 = 4th

Q_3 = 12th

From the pet group:

The data elements at the 4th and 12th positions are 68 and 79

So, we have:

Q_1 = 68

Q_3= 79

The lower and upper limits of the outlier are:

L = Q_1 - 1.5 \times (Q_3 - Q_1)

U = Q_3 + 1.5 \times (Q_3 - Q_1)

So, we have:

L = 68 - 1.5 \times (79 - 68)

L = 51.5

U = 79+ 1.5 \times (79 - 68)

U = 95.5

This means that data below 51.5 or above 95.5 are outliers.

<em>Hence, 99 is an outlier because 99 is greater than 95.5</em>

<u>(b) The parallel box plot</u>

The three groups are:

<em>Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99</em>

<em>Erlento: 88 80 80 81 92 87 88 81 82 80 87 92 87 80 82 </em>

<em>Alone: 62 70 73 75 77 80 84 84 84 87 87 87 90 91 99</em>

<em />

See attachment for the parallel box plots

Read more about box plots and outliers at:

brainly.com/question/14940764

5 0
3 years ago
Help me plssssssssssssssssss
meriva

Answer:

kmkioppppppppoiiu yw quip all cm ms all pure 100.00

3 0
2 years ago
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