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VLD [36.1K]
3 years ago
8

Find the sum of the interior angle measures of a convex 21-gon

Mathematics
1 answer:
Advocard [28]3 years ago
7 0

Answer:

  3420°

Step-by-step explanation:

The sum of angle measures of an n-gon is ...

  angle total = (n-2)×180°

For n = 21, the angle total is ...

  21-gon total = (21 -2)×180° = 3420°

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Mashcka [7]

Answer:

1)

M: 15

T: 14

W: 19

Th: 20

F: 9

Sa: 9

Su: 5

2) 18 degrees

Step-by-step explanation:

4 0
2 years ago
ABCD is a parallelogram. AB>AD. Prove: m∠ADB>m∠BDC. Show your work (statement reason maybe?)
Gwar [14]

In the parallelogram ABCD, join BD.

Consider the triangle Δ ABD.

It is given that AB > AD.

Since, in a triangle, angle opposite to longer side is larger, we have,

∠ ADB > ∠ ABD. --- (1)

Also, AB || DC and BD is a transversal.

Therefore,

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3 years ago
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Ivenika [448]

Answer

263/999                                                                                                                                                          

Step-by-step explanation:

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2 years ago
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Alexxandr [17]

Answer:

-1.375

Step-by-step explanation:

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2 years ago
A random sample of 16 one-kilogram sugar packets is obtained and the actual weights of the packets are measured. The sample mean
elena55 [62]

Answer:

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.905([tex]t_{995}). So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*0.058 = 0.171

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 1.053 - 0.171 = 0.882kg

The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.

The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.

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3 years ago
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