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crimeas [40]
2 years ago
8

I need help solving this difficult math problem step by step to get the answer. -10+-30/10 x 3

Mathematics
1 answer:
MAXImum [283]2 years ago
5 0

Answer:

The answer is -19

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Suppose that the data for analysis includes the attributeage. Theagevalues for the datatuples are (in increasing order) 13, 15,
Bas_tet [7]

Answer:

a) \bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96

Median = 25

b) Mode = 25, 35

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

c) Midrange = \frac{70+13}{3}=41.5

d) Q_1 = \frac{20+21}{2} =20.5

Q_3 =\frac{35+35}{2}=35

e) Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

f) Figura attached.

g) When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

Step-by-step explanation:

For this case w ehave the following dataset given:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

Part a

The mean is calculated with the following formula:

\bar X = \frac{\sum_{i=1}^{27} X_i }{27}= \frac{809}{27}=29.96

The median on this case since we have 27 observations and that represent an even number would be the 14 position in the dataset ordered and we got:

Median = 25

Part b

The mode is the most repeated value on the dataset on this case would be:

Mode = 25, 35

Since 25 and 35 are repeated 4 times, so then the distribution would be bimodal.

Part c

The midrange is defined as:

Midrange = \frac{Max+Min}{2}

And if we replace we got:

Midrange = \frac{70+13}{3}=41.5

Part d

For the first quartile we need to work with the first 14 observations

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25

And the Q1 would be the average between the position 7 and 8 from these values, and we got:

Q_1 = \frac{20+21}{2} =20.5

And for the third quartile Q3 we need to use the last 14 observations:

25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70

And the Q3 would be the average between the position 7 and 8 from these values, and we got:

Q_3 =\frac{35+35}{2}=35

Part e

The five number summary for this case are:

Min = 13 , Q1 = 20.5, Median=25, Q3= 35, Max = 70

Part f

For this case we can use the following R code:

> x<-c(13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30,33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70)

> boxplot(x,main="boxplot for the Data")

And the result is on the figure attached. We see that the dsitribution seems to be assymetric. Right skewed with the Median<Mean

Part g

When we use a quantile plot is because we want to show the percentage or the fraction of values below or equal to an specified value for the distribution of the data.

By the other hand the quantile-quantile plot shows the quantiles of the distribution values against other selected distribution (specified, for example the normal distribution). If the points are on a straight line we assume that the data values fit very well to the hypothetical distribution selected.

6 0
3 years ago
Only 4 b and c please
Artyom0805 [142]

Answer:

a) 2L + 2W = 42

b) L = 2W + 3

c) The system is :

2L + 2W = 42

L - 2W = 3

( add the two equations )

3L + (2w - 2w) = 45

3L = 45

L = 45/3

L = 15 ( Length )

L - 2W = 3

15 - 2W = 3

2W = 15 - 3

W = 12/2

W = 6

4 0
2 years ago
Prove that<br> 1+cosθ+sinθ/1+cosθ-sinθ = 1+sinθ/cosθ
marusya05 [52]

Step-by-step explanation:

(1 + cos θ + sin θ) / (1 + cos θ − sin θ)

Multiply by the reciprocal:

(1 + cos θ + sin θ) / (1 + cos θ − sin θ) × (1 + cos θ + sin θ) / (1 + cos θ + sin θ)

(1 + cos θ + sin θ)² / [ (1 + cos θ − sin θ) (1 + cos θ + sin θ) ]

(1 + cos θ + sin θ)² / [ (1 + cos θ)² − sin² θ) ]

Distribute and simplify:

(1 + cos θ + sin θ)² / (1 + 2 cos θ + cos² θ − sin² θ)

[ 1 + 2 (cos θ + sin θ) + (cos θ + sin θ)² ] / (1 + 2 cos θ + cos² θ − sin² θ)

(1 + 2 cos θ + 2 sin θ + cos² θ + 2 sin θ cos θ + sin² θ) / (1 + 2 cos θ + cos² θ − sin² θ)

Use Pythagorean identity:

(2 + 2 cos θ + 2 sin θ + 2 sin θ cos θ) / (sin² θ + cos² θ + 2 cos θ + cos² θ − sin² θ)

(2 + 2 cos θ + 2 sin θ + 2 sin θ cos θ) / (2 cos² θ + 2 cos θ)

(1 + cos θ + sin θ + sin θ cos θ) / (cos² θ + cos θ)

Factor:

(1 + cos θ + sin θ (1 + cos θ)) / (cos θ (1 + cos θ))

(1 + cos θ)(1 + sin θ) / (cos θ (1 + cos θ))

(1 + sin θ) / cos θ

3 0
3 years ago
What is equivalent to 32
frez [133]

Answer:

32.0 or 32/1 or 64/2

4 0
2 years ago
Read 2 more answers
Which interval is the solution set to 0.35x - 4.8 &lt; 5.2 -0.9x?
8_murik_8 [283]

Answer:  (-\infty, 8)

Step-by-step explanation:

Given the following inequality:

0.35x - 4.8 < 5.2 -0.9x

You need to solve for "x" in order to find the solution.

The steps are:

1. Add 0.9x to both sides of the inequality:

0.35x - 4.8+0.9x < 5.2 -0.9x+0.9x\\\\1.25x-4.8

2. Add 4.8 to both sides:

1.25x-4.8+4.8

3. Divide both sides by 1.25:

\frac{1.25x}{1.25}

Notice that "x" is  less than 8. This indicates that 8 is not included in the solution and you must use parentheses.

The solution in Interval notation is:

(-\infty, 8)

8 0
3 years ago
Read 2 more answers
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