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Alex787 [66]
3 years ago
10

Is this linear or non-linear??

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
7 0

Answer:

None-Linear

Step-by-step explanation:

Because there is no pattern from one Y to the next

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Here is a list of the normal monthly precipitation (in inches) for January for 20 different U.S. cities. 4.0, 1.0, 1.5, 1.6, 2.0
shutvik [7]

Answer:

minimum value = 1.0 , the lower quartile = 2.3, the middle value = 3.5.

the upper quartile = 3.95, the maximum value  = 7.3.

Step-by-step explanation:

Five Number Summary of any data consists basically of 5 data sets.

1) The minimum Value

2)The first (lower) quartile

3) The middle value (Median)

4) The third (upper) quartile

5)The maximum value

Now, the given data is  4.0, 1.0, 1.5, 1.6, 2.0, 2.2, 2.4, 2.7, 3.4, 3.4, 3.5, 3.6, 3.6, 3.7, 3.7, 3.9, 4.1, 5.8, 4.1, 7.3

Arrange it in the ascending order,

1.0, 1.5, 1.6, 2.0, 2.2, 2.4, 2.7, 3.4, 3.4, 3.5, 3.6, 3.6, 3.7, 3.7, 3.9, 4.0, 4.1, 4.1, 5.8, 7.3

Here,

1) The minimum Value  is 1.0

2)The first (lower) quartile  is  2.3.

3) The middle value (Median)  is 3.5.

4) The third (upper) quartile is  3.95.

5)The maximum value  is 7.3.

7 0
3 years ago
Please help on the area for 7 8 9 please all i need help is
weeeeeb [17]

The area of the equilateral, isosceles and right angled triangle are 12.6mm², 9.61in² and 16.81yds² respectively.

<h3>What is the area of the equilateral, isosceles and right angle triangle?</h3>

Note that:

The area of an Equilateral triangle is expressed as A = ((√3)/4)a²

Where a is the dimension of the side.

The area of an Isosceles triangle is expressed as A = (ah)/2

Where a is the dimension of the base and h is the height.

The area of a Right angled triangle is expressed as A = (ab)/2

Where a and b is the dimension of the two sides other than the hypotenuse.

For the Equilateral triangle.

Given that;

  • a = 5.4mm
  • Area A = ?

A = ((√3)/4)(5.4mm)²

A = ((√3)/4)( 29.16mm² )

A = 12.6mm²

Area of the Equilateral triangle is 12.6mm²

For the Isosceles triangle.

Given that;

  • Base a = 3.4in
  • Slant height b = 5.9in
  • height h = ?
  • Area A = ?

The height h is the imaginary line drawn upward from the center of a.

First, we calculate the height using Pythagorean theorem

x² = y² + z²

Where x = b = 5.9in, y = a/2 = 3.4in/2 = 1.7in, and z = h

(5.9in)² = (1.7in)² + h²

34.81in² = 2.89in² + h²

h² = 34.81in² - 2.89in²

h² = 31.92in²

h = √31.92in²

h = 5.65in

Now, the area will be;

A = (ah)/2

A = (3.4in × 5.65in )/2

A = 19.21in²/2

A = 9.61in²

Area of the Isosceles triangle is 9.61in².

For the Right angled triangle

Given that;

  • a = 8.2yds
  • b = 4.1yds
  • c = 9.17yds
  • Area A = ?

A = (ab)/2

A = ( 8.2yds × 4.1yds)/2

A = ( 33.62yds²)/2

A = 16.81yds²

Area of the Right angled triangle is 16.81yds²

Therefore, the area of the equilateral, isosceles and right angled triangle are 12.6mm², 9.61in² and 16.81yds² respectively.

Learn more about Pythagorean theorem here: brainly.com/question/343682

#SPJ1

6 0
2 years ago
Given the probability distributions shown to the​ right, complete the following parts.
Elan Coil [88]

Answer:

a) Expected Value for distribution A, E(X) = 3.020

Expected Value for distribution B, E(X) = 0.980

b) Standard deviation of distribution A = 1.157

Standard deviation of distribution B = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

Step-by-step explanation:

Expected values is given as

E(X) = Σ xᵢpᵢ

where xᵢ = each possible sample space

pᵢ = P(X=xᵢ) = probability of each sample space occurring.

Distributions A and B is given by

X P(X) X P(x)

0 0.04 0 0.47

1 0.09 1 0.25

2 0.15 2 0.15

3 0.25 3 0.09

4 0.47 4 0.04

For distribution A

E(X) = Σ xᵢpᵢ = (0×0.04) + (1×0.09) + (2×0.15) + (3×0.25) + (4×0.47) = 3.02

For distribution B

E(X) = Σ xᵢpᵢ = (0×0.47) + (1×0.25) + (2×0.15) + (3×0.09) + (4×0.04) = 0.98

b) Standard deviation = √(variance)

But Variance is given by

Variance = Var(X) = Σx²p − μ²

where μ = E(X)

For distribution A

Σx²p = (0²×0.04) + (1²×0.09) + (2²×0.15) + (3²×0.25) + (4²×0.47) = 10.46

Variance = Var(X) = 10.46 - 3.02² = 1.3396

Standard deviation = √(1.3396) = 1.157

For distribution B

Σx²p = (0²×0.47) + (1²×0.25) + (2²×0.15) + (3²×0.09) + (4²×0.04) = 2.30

Variance = Var(X) = 2.30 - 0.98² = 1.3396

Standard deviation = √(1.3396) = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

7 0
3 years ago
Nine more than the quotient of two and a number x.
Ostrovityanka [42]

Answer (2÷x) +9

Step-by-step explanation:

4 0
3 years ago
Domain of (x-6)/(square root of x^2+2x-15)
ki77a [65]

Answer:

(X-6) √x^2+2x-15 \ x^2 +2x - 15

5 0
2 years ago
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