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3 years ago

What is the midpoint of the line segment graphed below?

Mathematics

2 answers:

Diano4ka-milaya [45]3 years ago

6 0

I’m pretty sure it’s b because
You find your midpoint then you would divide after that I’m pretty sure you round

dybincka [34]3 years ago

4 0

Answer:

Step-by-step explanation:

Use the midpoint calculator to find out the midpoint of a line segment, ... A graph showing how to find the midpoint of a segment on the Cartesian plane ... As a supplement to this calculator, we have written an article below that ... Divide 10 by 2, the result of which is 5, this is the y coordinate of the midpoint.

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WILL GIVE BRAINLY PLEASE HELP!!!!!!!

marta [7]

Answer:

r² = 0.5652 < 0.7 therefore, the correlation between the variables does not imply causation

Step-by-step explanation:

The data points are;

X, Y

0.7, 1.11

21.9, 3.69

18, 4

16.7, 3.21

18, 3.7

13.8, 1.42

18, 4

13.8, 1.42

15.5, 3.92

16.7, 3.21

The correlation between the values is given by the relation

Y = b·X + a

$b = \dfrac{N\sum XY - \left (\sum X \right )\left (\sum Y \right )}{N\sum X^{2} - \left (\sum X \right )^{2}}$

$a = \dfrac{\sum Y - b\sum X}{N}$

Where;

N = 10

∑XY = 499.354

∑X = 153.1

∑Y = 29.68

∑Y² = 100.546

∑X² = 2631.01

(∑ X)² = 23439.6

(∑ Y)² = 880.902

From which we have;

$b = \dfrac{10 \times 499.354 -153.1 \times 29.68}{10 \times 2631.01 - 23439.6} = 0.1566$

$a = \dfrac{29.68 - 0.1566 \times 153.1}{10} = 0.5704$

$r = \dfrac{N\sum XY - \left (\sum X \right )\left (\sum Y \right )}{\sqrt{\left [N\sum X^{2} - \left (\sum X \right )^{2} \right ]\times \left [N\sum Y^{2} - \left (\sum Y \right )^{2} \right ]}}$

$r = \dfrac{10 \times 499.354 -153.1 \times 29.68}{\sqrt{\left (10 \times 2631.01 - 23439.6 \right )\times \left (10 \times 100.546- 880.902\right )} } = 0.7518$

r² = 0.5652 which is less than 0.7 therefore, there is a weak relationship between the variables, and it does not imply causation.

7 0

3 years ago

A running circuit is in the shape of a triangle with lengths of 6km, 6.5km and 7km. What are the sizes of the angles (in minutes

Rudik [331]

A triangle is an example of a class of figures referred to as plane shapes. It has three straight sides and three internal angles which sum up to $180^{o}$ . The measures of the internal angles of the triangle given in the question are A = $52.6^{o}$ , B = $59.4^{o}$ , and C = $68^{o}$ .

Considering the given question, let the sides of the triangle be: a = 6 km, b = 6.5 km, and c = 7 km.

Apply the Cosine rule to have:

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab Cos C

So that;

$7^{2}$ = $6^{2}$ + $(6.5)^{2}$ - 2(6 * 6.5) Cos C

49 = 36 + 42.25 - 78Cos C

78 Cos C = 78.25 - 49

= 29.25

Cos C = $\frac{29.25}{78}$

= 0.375

C = $Cos^{-1}$ 0.375

= 67.9757

C = $68^{o}$

Apply the Sine rule to determine the value of B,

$\frac{b}{Sin B}$ = $\frac{c}{Sin C}$

$\frac{6.5}{Sin B}$ = $\frac{7}{Sin 68}$

SIn B = $\frac{6.5 *Sin 68}{7}$

= 0.861

B = $Sin^{-1}$ 0.861

= 59.43

B = $59.4^{o}$

Thus to determine the value of A, we have;

A + B + C = $180^{o}$

A + $59.4^{o}$ + $68^{o}$ = $180^{o}$

A = $180^{o}$ - 127.4

= 52.6

A = $52.6^{o}$

Therefore the sizes of the internal angles of the triangle are: A = $52.6^{o}$ , B = $59.4^{o}$ , and C = $68^{o}$ .

For more clarifications on applications of the Sine and Cosine rules, visit: brainly.com/question/14660814

#SPJ1

8 0

2 years ago

Describe the location of the point having the following coordinates.

Maslowich

Answer:

Step-by-step explanation:

Quadrant II or Quadrant IV

Quadrant I or Quadrant II

Quadrant I or Quadrant IV

Quadrant III or Quadrant IV

6 0

3 years ago

Suppose that a random sample of size 25 is to be selected from a population with mean 41 and standard deviation 9. What is the a

lorasvet [3.4K]

Answer:

The probability that X will be more than 0.5 away fro the population

mean is 0.7812 ⇒ answer a

Step-by-step explanation:

* Let us explain how to solve the problem

- The sample size n is 25

- The mean of the population μ is 41

- The standard deviation σ is 9

- We need to find the approximate probability that X will be more than

0.5 away from the population mean

∵ z-score = (X - μ)/(σ/√n)

∵ X - μ = 0.5

- That means X > 41 + 0.5, OR X < 41 - 0.5

- Then we need to find P(X > 41.5) and P(X < 40.5)

∵ z = $\frac{41.5-41}{\frac{9}{\sqrt{25}}}$ = 0.2778

∵ z = $\frac{40.5-41}{\frac{9}{\sqrt{25}}}$ = -0.2778

- Let us use the normal distribution table to find the corresponding

area to z-score

∵ P(z < 0.2778) = 0.6116

∵ P(z > -0.2778) = 0.3928

- P(40.5 < X < 41.5) = P( -0.2778 < z < 0.2778)

∴ The probability of X to be with in 0.5 = 0.6116 - 0.3928

∴ The probability of X to be with in 0.5 = 0.2188

∵ The probability of X to be more than 0.5 away from the population

mean = 1 - 0.2188

∴ P(X > 41.5) OR P(X < 40.5) = 0.7812

* The probability that X will be more than 0.5 away fro the population

mean is 0.7812

8 0

3 years ago

GIVING BRAINLIEST FOR BEST ANSWER!

Setler79 [48]

7 / 7 = 1

3 + 1 = 4

4 + 3 and 1 / 3 = 7 and 1 / 3

6 0

2 years ago

Read 2 more answers