18.09.2020

What is a equal or greater two

. 4

Faq

Mathematics
Step-by-step answer
P Answered by PhD

344=12x+80

a=344

b=12

c=80

Step-by-step explanation:

Let x be number of balls shot through the hoop by Central City High School's robot.

We have been given an equation a=bx+c and we are asked to find the values of a, b and c.

We are told that Central City High School's team earned an additional 12 points for each ball the robot shot through a hoop. So the points earned by shooting x balls through the hoop will be 12x.

The team also earned 80 points for their robot climbing over an obstacle. So total number of points earned by team will be equal to 12x+80.

We are also told that the team won the round with 344 total points. We can represent this information in an equation as: 344=12x+80

Upon comparing this equation with our given equation we will get,

a=344

b=12

c=80

Therefore, the equation 344=12x+80 can be used to find the number of balls, x, team's robot shot through the hoop.

Mathematics
Step-by-step answer
P Answered by PhD

344=12x+80

a=344

b=12

c=80

Step-by-step explanation:

Let x be number of balls shot through the hoop by Central City High School's robot.

We have been given an equation a=bx+c and we are asked to find the values of a, b and c.

We are told that Central City High School's team earned an additional 12 points for each ball the robot shot through a hoop. So the points earned by shooting x balls through the hoop will be 12x.

The team also earned 80 points for their robot climbing over an obstacle. So total number of points earned by team will be equal to 12x+80.

We are also told that the team won the round with 344 total points. We can represent this information in an equation as: 344=12x+80

Upon comparing this equation with our given equation we will get,

a=344

b=12

c=80

Therefore, the equation 344=12x+80 can be used to find the number of balls, x, team's robot shot through the hoop.

Mathematics
Step-by-step answer
P Answered by Specialist
The blue line is the solution.

That full blue circle means "equal" and the blue line is between -4 and 3 so the solution is x greather than or equal to -4 AND x less than or equal to 3. You can write it as - 4 ≤ x ≤ 3 (if you read it, -4 is less than x that is less than 3, solutions are between this range, so -4, -3, -2, -1, 0, 1, 2, 3)

Hope it's right and hope it's helpful!
Mathematics
Step-by-step answer
P Answered by Specialist
The blue line is the solution.

That full blue circle means "equal" and the blue line is between -4 and 3 so the solution is x greather than or equal to -4 AND x less than or equal to 3. You can write it as - 4 ≤ x ≤ 3 (if you read it, -4 is less than x that is less than 3, solutions are between this range, so -4, -3, -2, -1, 0, 1, 2, 3)

Hope it's right and hope it's helpful!
Mathematics
Step-by-step answer
P Answered by PhD

(1) Less than → <

(2) Greater than → >

(3) Less than or equal to → ≤

(4) Greater than or equal to → ≥

(5) At most → ≤

(6) At least → ≥

(7) Maximum → ≤

(8) Minimum → ≥

(9) No more than → \ngtr

(10) No less than → \nless

(11) Fewer than → <

(12) More than → >

Step-by-step explanation:

Here we have to type the symbols for the written conditions given in the problem.

(1) Less than → <

(2) Greater than → >

(3) Less than or equal to → ≤

(4) Greater than or equal to → ≥

(5) At most → ≤

(6) At least → ≥

(7) Maximum → ≤

(8) Minimum → ≥

(9) No more than → \ngtr

(10) No less than → \nless

(11) Fewer than → <

(12) More than → >

Mathematics
Step-by-step answer
P Answered by PhD

(1) Less than → <

(2) Greater than → >

(3) Less than or equal to → ≤

(4) Greater than or equal to → ≥

(5) At most → ≤

(6) At least → ≥

(7) Maximum → ≤

(8) Minimum → ≥

(9) No more than → \ngtr

(10) No less than → \nless

(11) Fewer than → <

(12) More than → >

Step-by-step explanation:

Here we have to type the symbols for the written conditions given in the problem.

(1) Less than → <

(2) Greater than → >

(3) Less than or equal to → ≤

(4) Greater than or equal to → ≥

(5) At most → ≤

(6) At least → ≥

(7) Maximum → ≤

(8) Minimum → ≥

(9) No more than → \ngtr

(10) No less than → \nless

(11) Fewer than → <

(12) More than → >

Mathematics
Step-by-step answer
P Answered by PhD

Explained below.

Step-by-step explanation:

The question is:

Compare the distributions using either the means and standard deviations or the five-number summaries. Justify your choice.

Set A: {36, 51, 37, 42, 54, 39, 53, 42, 46, 38, 50, 47}

Set B: {22, 57, 46, 24, 31, 41, 64, 50, 28, 59, 65, 38}

The five-number summary is:

MinimumFirst Quartile Median Third Quartile Maximum

The five-number summary for set A is:

Variable   Minimum       Q₁         Median        Q₃         Maximum

  Set A       36.00       38.25        44.00       50.75        54.00

The five-number summary for set B is:

Variable   Minimum       Q₁         Median        Q₃         Maximum

 Set B         22.00      28.75        48.00       58.50         65.00

Compute the mean for both the data as follows:

Mean_{A}=\frac{1}{12}\times [36+51+37+...+47]=44.58\approx 44.6\\\\Mean_{B}=\frac{1}{12}\times [22+57+46+...+38]=44.58\approx 44.6

Both the distribution has the same mean.

Compare mean and median for the two data:

Mean_{A}Median_{A}\\\\Mean_{B}Median_{B}

This implies that set A is positively skewed whereas set B is negatively skewed.

Compute the standard deviation for both the set as follows:

SD_{A}=\sqrt{\frac{1}{12-1}\times [(36-44.6)^{2}+...+(47-44.6)^{2}]}=6.44\approx 6.4\\\\SD_{B}=\sqrt{\frac{1}{12-1}\times [(22-44.6)^{2}+...+(38-44.6)^{2}]}=15.56\approx 15.6

The set B has a greater standard deviation that set A. Implying set B has a greater variability that set B.
Mathematics
Step-by-step answer
P Answered by Master

A type II error is failing to reject the hypothesis that μ is equal to 2800 when in fact, μ is less than 2800.

Step-by-step explanation:

A Type II error happens when a false null hypothesis is failed to be rejected.

The outcome (the sample) probability is still above the level of significance, so it is consider that the result can be due to chance (given that the null hypothesis is true) and there is no enough evidence to claim that the null hypothesis is false.

In this contest, a Type II error would be not rejecting the hypothesis that the mean lifetime of the light bulbs is 2800 hours, when in fact this is false: the mean lifetime is significantly lower than 2800 hours.

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