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3 years ago

Please help!

Mathematics

1 answer:

insens350 [35]3 years ago

3 0

Answer:

2a. 5.8

2b. On average, a score will be 5.8 points away from the average score.

2c. The score will be about 2 points nearer to the average.

4a1. Sample 1 Mean: $2300; Sample 1 Median: $2250.

4a2. Sample 2 Mean: $5650; Sample 2 Median: $5600.

4b. The monthly cost of living is more expensive for a married couple w/ 1 child than for a single person.

Step-by-step explanation:

2a./2b.

In order to find the mean absolute deviation, we need to first find the mean.

(21 + 45 + 21 + 14 + 21 + 28 + 24 + 14 + 24 + 28)/10

(66 + 35 + 49 + 38 + 52)/10

(101 + 87 + 52)/10

(101 + 139)/10

240/10

The mean of this set is 24.

Now for the deviation.

21 - 24 = -3
45 - 24 = 21
14 - 24 = -10
28 - 24 = 4
24 - 24 = 0

For the mean absolute deviation, divide the combined value of the 10 deviation values from before (in absolute value) by 10.

(|-3| + 21 + |-3| + |-10| + |-3| + 4 + |-10| + 4)/10

(3 + 25 + 3 + 10 + 3 + 10 + 4)/10

(28 + 13 + 13 + 4)/10

(41 + 17)/10

58/10

The mean absolute deviation is 5.8. That interpreted would be like this: A score, on average, is 5.8 points away from the average score.

2c.

What I'll do here is the following: subtract 45 from 240, and then subtract 21 from the mean deviations. In both cases, I will now divide by 9.

(240 - 45)/9

195/9

65/3

The mean of the set (without 45) is now 21 2/3.

(58 - 21)/9

37/9

The new mean absolute deviation is 4 1/9, which rounds to 4.1. That's 1.7 points nearer to the average than originally.

So, without 45 as one of the scores, the M.A.D. decreases by about 2. That interpreted would be like this: The scores are about 2 points nearer to the average than before.

4a./4b.

First, I'll find the mean for sample one.

(1800 + 2000 + 2200 + 2300 + 2500 + 3000)/6

(3800 + 4500 + 5500)/6

(8300 + 5500)/6

13800/6

The mean for sample one is $2,300/month.

Now for the median for sample one. Since there are 6 values, add the two middle values and divide by 2.

(2200 + 2300)/2

4500/2

The median for sample one is $2,250/month.

Now for the mean & median for sample two.

Mean:

(5400 + 5500 + 5600 + 5600 + 5800 + 6000)/6

(10900 + 11200 + 11800)/6

(22100 + 11800)/6

33900/6

The mean for sample two is $5,650/month.

Median:

Again, there are 6 values. However, this time they are the same, so there is no need to divide. The median for sample two is $5,600/month.

Looking at the data, the mean & median for sample was larger than sample one. Thus, I can conclude the following: The monthly cost of living is more expensive for a married couple w/ 1 child than for a single person.

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Answer:

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Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

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Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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