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2 years ago

Which scatter plot has a correlation coefficient closest to r=1

Mathematics

2 answers:

AleksAgata [21]2 years ago

7 0

Answer:

Step-by-step explanation:

E2020

Alex73 [517]2 years ago

5 0

Answer:

The correct answer is the scatter plot which has a strong positive slope of the curve of best fit.

Step-by-step explanation:

A scatter plot shows a relationship between two variables for a particular individual. It is actually a graph where we draw a curve of best fit which can represent the whole data. A scatter plot having the correlation coefficient closest to r = 1, is said to have a very strong correlation between the variables.

A scatter plot would have a strong correlation if the data points of the individual are very close to the line or curve of best fit.

For r = 1 , the correlation is strong and positive.

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Jordan solved the equation −7x + 25 = 48; his work is shown below. Identify the error and where it was made.

Mashcka [7]

The answer is C, Step 3: He should have divided both sides by -7.

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

4 0

2 years ago

The amount of time it takes for a student to complete a statistics quiz is uniformly distributed (or, given by a random variable

topjm [15]

Answer:

(A) 0.15625

(B) 0.1875

(C) Can't be computed

Step-by-step explanation:

We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.

Let X = Amount of time taken by student to complete a statistics quiz

So, X ~ U(32 , 64)

The PDF of uniform distribution is given by;

f(X) = $\frac{1}{b-a}$ , a < X < b where a = 32 and b = 64

The CDF of Uniform distribution is P(X <= x) = $\frac{x-a}{b-a}$

(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)

P(X > 59) = 1 - P(X <= 59) = 1 - $\frac{x-a}{b-a}$ = 1 - $\frac{59-32}{64-32}$ = $1-\frac{27}{32}$ = 0.15625

(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)

P(X <= 43) = $\frac{43-32}{64-32}$ = $\frac{11}{32}$ = 0.34375

P(X < 37) = $\frac{37-32}{64-32}$ = $\frac{5}{32}$ = 0.15625

P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875

= P(X = 44.74)

The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.

3 0

2 years ago

Which steps can be used to solve StartFraction 6 Over 7 EndFraction x + one-half = StartFraction 7 Over 8 EndFraction for x? Che

34kurt

Answer:

Subtract One-half from both sides of the equation.

Divide both sides by 6/7

Multiply both sides by 7/6

Step-by-step explanation:

just took the question on ed

8 0

2 years ago

Read 2 more answers

Shawn purchased $87.06 worth of merchandise at a department store. He gave the clerk $100. Round to the nearest whole dollar to

Alborosie

Answer:

it should be 187.6 if u got it wrong let me know so i can give u another answer

Step-by-step explanation:

9 0

2 years ago

Find the distance from (4, −7, 6) to each of the following.

LenKa [72]

Answer:

(a) 6 units

(b) 4 units

(d) 9.22 units

(e) 7.21 units

(f) 8.06 units

Step-by-step explanation:

The distance d from one point (x₁, y₁, z₁) to another point (x₂, y₂, z₂) is given by;

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Now from the question;

(a) The distance from (4, -7, 6) to the xy-plane

The xy-plane is the point where z is 0. i.e

xy-plane = (4, -7, 0).

Therefore, the distance d is from (4, -7, 6) to (4, -7, 0)

d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]

d = √[(0)² + (0)² + (-6)²]

d = √(-6)²

d = √36

d = 6

Hence, the distance to the xy plane is 6 units

(b) The distance from (4, -7, 6) to the yz-plane

The yz-plane is the point where x is 0. i.e

yz-plane = (0, -7, 6).

Therefore, the distance d is from (4, -7, 6) to (0, -7, 6)

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]

d = √[(4)² + (0)² + (0)²]

d = √(4)²

d = √16

d = 4

Hence, the distance to the yz plane is 4 units

(c) The distance from (4, -7, 6) to the xz-plane

The xz-plane is the point where y is 0. i.e

xz-plane = (4, 0, 6).

Therefore, the distance d is from (4, -7, 6) to (4, 0, 6)

d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]

d = √[(0)² + (-7)² + (0)²]

d = √[(-7)²]

d = √49

d = 7

Hence, the distance to the xz plane is 7 units

(d) The distance from (4, -7, 6) to the x axis

The x axis is the point where y and z are 0. i.e

x-axis = (4, 0, 0).

Therefore, the distance d is from (4, -7, 6) to (4, 0, 0)

d = √[(4 - 4)² + (-7 - 0)² + (6 - 0)²]

d = √[(0)² + (-7)² + (6)²]

d = √[(-7)² + (6)²]

d = √[(49 + 36)]

d = √(85)

d = 9.22

Hence, the distance to the x axis is 9.22 units

(e) The distance from (4, -7, 6) to the y axis

The x axis is the point where x and z are 0. i.e

y-axis = (0, -7, 0).

Therefore, the distance d is from (4, -7, 6) to (0, -7, 0)

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 0)²]

d = √[(4)² + (0)² + (6)²]

d = √[(4)² + (6)²]

d = √[(16 + 36)]

d = √(52)

d = 7.22

Hence, the distance to the y axis is 7.21 units

(f) The distance from (4, -7, 6) to the z axis

The z axis is the point where x and y are 0. i.e

z-axis = (0, 0, 6).

Therefore, the distance d is from (4, -7, 6) to (0, 0 6)

d = √[(4 - 0)² + (-7 - (0))² + (6 - 6)²]

d = √[(4)² + (-7)² + (0)²]

d = √[(4)² + (-7)²]

d = √[(16 + 49)]

d = √(65)

d = 8.06

Hence, the distance to the z axis is 8.06 units

5 0

2 years ago