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Olin [163]
2 years ago
6

The size of angle aob is equal to 132 degrees and the size of angle cod is equal to 141 degrees. find the size of angle dob.

Mathematics
1 answer:
Varvara68 [4.7K]2 years ago
8 0

angle AOB = 132 and is also the sum of angles AOD and DOB. Hence 
angle AOD + angle DOB = 132°    ---> 1 

angle COD = 141 and is also the sum of angles COB and BOD. Hence 
angle COB + angle DOB = 141°    ---> 2

Now we add the left sides together and the right sides of equations 1 and 2 together to form a new equation. 

angle AOD + angle DOB + angle COB + angle DOB = 132 + 141       ---> 3 

We should also note that: 
angle AOD + angle DOB + angle COB = 180° 

Therefore substituting angle AOD + angle DOB + angle COB in equation 3 by 180 and solving for angle DOB:
180 + angle DOB = 132 + 141 
angle DOB = 273 - 180 = 93° 

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Answer:

(a) The variance decreases.

(b) The variance increases.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the sample mean is given by,

\mu_{\bar x}=\mu

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\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The standard deviation of sample mean is inversely proportional to the sample size, <em>n</em>.

So, if <em>n</em> increases then the standard deviation will decrease and vice-versa.

(a)

The sample size is increased from 64 to 196.

As mentioned above, if the sample size is increased then the standard deviation will decrease.

So, on increasing the value of <em>n</em> from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for <em>n</em> = 64 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7

The standard deviation of the sample mean for <em>n</em> = 196 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4

The standard deviation of the sample mean decreased from 0.7 to 0.4 when <em>n</em> is increased from 64 to 196.

Hence, the variance also decreases.

(b)

If the sample size is decreased then the standard deviation will increase.

So, on decreasing the value of <em>n</em> from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for <em>n</em> = 784 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2

The standard deviation of the sample mean for <em>n</em> = 49 is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8

The standard deviation of the sample mean increased from 0.2 to 0.8 when <em>n</em> is decreased from 784 to 49.

Hence, the variance also increases.

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2 years ago
Your company has offices in 2 cities. The Denver office has 137 employees with an average salary of $67,013. The Anchorage offic
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Alright, lets get started.

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So the total salay of all employees at Denver office will be = 137*67013

So the total salay of all employees at Denver office will be = $ 9180781

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So, complete total salary of employees at both offices will be = 9180781+1783300

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