Use a special right triangle to write cos 30

The angles are within the range of angles are 120°, 110°, 130°  and  120°, 120°, 120°

From the given options, the angles are within the range of angles are 120°, 110°, 130°  and  120°, 120°, 120°

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The question is about work and efficiency and requires solving a system of equations to find the individual work rates of people from different nationalities. However, because we have insufficent data (two equations but three unknowns), we cannot find a unique solution and thus can't determine how long it would take for one Chinese, one Indian, and one Japanese to complete the work together.

The student's schoolwork question involves solving a problem related to work and efficiency, which is a part of mathematics, specifically related to rate and work problems.

First, let's find the rate of work done by individuals from each nationality. We can set up the equations based on the given information:

We will need to solve this system of equations to find the individual efficiencies I, C, and J. However, we have two equations and three unknowns, so we cannot find unique solutions for I, C, and J unless we have one more equation or information.
Since we cannot determine the individual efficiencies with the given information, we cannot determine how long it would take one Chinese, one Indian, and one Japanese to complete the work.

Answer: its  410439

(all u need to do is subtract!)

Answer its,    C and standard deviation,  lowest  

Step-by-step explanation:

Answer:

Option A The slope of the line is [tex]-5[/tex]

Step-by-step explanation:

we know that

The equation of the line into slope intercept form is equal to

[tex]y=mx+b[/tex]

where

m is the slope

b is the y-intercept

we have

[tex]y=-\frac{2}{3}-5x[/tex]

Rewrite

[tex]y=-5x-\frac{2}{3}[/tex]

therefore

[tex]b=-\frac{2}{3}[/tex] -----> the y-intercept

[tex]m=-5[/tex] ------> slope

The correct answer is: 5 over 15 multiplied by 4 over 14

Probability is a number that expresses the likelihood or chance that a specific event will take place. Both proportions ranging from 0 to 1 and percentages ranging from 0% to 100% can be used to describe probabilities.

Given this, Deja draws a marble from the bag randomly without looking.

The total number of marbles in the bag is:

5 + 8 + 2 = 15

Deja draws a marble from the bag randomly without looking. Therefore, the probability of drawing a red marble in the first trial is:

P(red on first draw) = 5/15

Deja draws another marble from the bag without replacing the first one. Therefore,

The probability of drawing a red marble in the second trial, given that she drew a red marble in the first trial, is:

P(red on second draw | red on first draw) = 4/14

We multiply these probabilities because the events are independent (i.e., the outcome of the first draw does not affect the outcome of the second draw).

Therefore, the expression that shows the probability of drawing red marbles in both trials is:

= P(red on the first draw) × P(red on second draw | red on the first draw)

= (5/15) × (4/14) = 4/63

So the correct answer is: 5 over 15 multiplied by 4 over 14.

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True, A set of data can indeed have no mode. In statistics, the mode represents the value(s) that appear most frequently in a dataset.

However, there are scenarios where no value occurs more frequently than others, resulting in a lack of a mode. This can happen in several situations. For example, in a uniform distribution where all values occur with equal frequency, or in a dataset with no repetition of values (each value occurs only once). Additionally, if the data is continuous and does not have distinct categories or intervals, it may not have a mode. In summary, while a mode is a common descriptive statistic, it's not a requirement for every dataset, and some datasets do indeed have no mode.

The linear equality represented by the graphis [tex]\boxed{{\mathbf{y > }}\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{x + 3}}}[/tex] and it matches with [tex]\boxed{{\mathbf{OPTION C}}}[/tex].

Further explanation:

From given graph, theline passes through points [tex]\left({3,5}\right)[/tex]  and [tex]\left({ - 3,1}\right)[/tex] as shown below in Figure 1.

The slope of a line passes through points [tex]\left({{x_1},{y_1}}\right)[/tex] and [tex]\left({{x_2},{y_2}}\right)[/tex] is calculated as follows:

[tex]m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}{\text{}}[/tex]   ......(1)

Here, the slope of a line is denoted as  and points are [tex]\left({{x_1},{y_1}}\right)[/tex] and [tex]\left({{x_2},{y_2}}\right)[/tex].

Substitute [tex]3[/tex] for [tex]{x_1}[/tex] , [tex]5[/tex] for [tex]{y_1}[/tex] , [tex]-3[/tex] for [tex]{x_2}[/tex] and [tex]1[/tex] for [tex]{y_2}[/tex] in equation (1) to obtain the slope of a line that passes through points [tex]\left( {3,5}\right)[/tex] and [tex]\left({ - 3,1}\right)[/tex].

[tex]\begin{aligned}m&=\frac{{1 - 5}}{{ - 3 - 3}}\\&=\frac{{ - 4}}{{ - 6}}\\&=\frac{2}{3}\\\end{aligned}[/tex]

Therefore, the slope is [tex]\frac{2}{3}[/tex].

The point-slope form of the equation of a line with slope   passes through point  is represented as follows:

[tex]y - {y_1} = m\left({x - {x_1}}\right){\text{}}[/tex]         ......(2)

Substitute [tex]3[/tex] for [tex]{x_1}[/tex] , [tex]5[/tex] for [tex]{y_1}[/tex] and [tex]\frac{2}{3}[/tex] for [tex]m[/tex] in equation (2) to obtain the equation of line.

[tex]\begin{aligned}y - 5 = \frac{2}{3}\left({x - 3} \right)\\3\left( {y - 5} \right) = 2x - 6\\3y - 15 = 2x - 6\\3y = 2x - 6 + 15\\\end{aligned}[/tex]

Further simplify the above equation.

[tex]\begin{aligned}3y=2x + 9\\y=\frac{2}{3}x+\frac{9}{3}\\y=\frac{2}{3}x + 3\\\end{aligned}[/tex]

Therefore, the value of [tex]y[/tex] is [tex]\dfrac{2}{3}x + 3[/tex].

Since the shaded part in Figure 1 is above the equation of line [tex]y = \dfrac{2}{3}x + 3[/tex], therefore, greater than sign is used instead of is equal to

Therefore, the inequality is [tex]y > \dfrac{2}{3}x + 3[/tex].

Now, the four options are given below.

[tex]\begin{aligned}{\text{OPTION A}}\to y < \frac{2}{3}x + 3 \hfill \\{\text{OPTION B}} \to y > \frac{3}{2}x + 3 \hfill \\{\text{OPTION C}} \to y > \frac{2}{3}x + 3 \hfill \\{\text{OPTION D}}\to y < \frac{3}{2}x + 3 \hfill\\\end{aligned}[/tex]

Since OPTION C matches the obtained equation that is [tex]y > \dfrac{2}{3}x + 3[/tex].

Thus, the linear equality represented by the graph is [tex]\boxed{{\mathbf{y > }}\frac{{\mathbf{2}}}{{\mathbf{3}}}{\mathbf{x + 3}}}[/tex]and it matches with [tex]\boxed{{\mathbf{OPTION C}}}[/tex]

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Answer Details:

Grade: Junior High School

Subject: Mathematics

Chapter: Coordinate Geometry

Keywords: Coordinate Geometry, linear equation, system of linear equations in two variables, variables, mathematics,equation of line, line, passes through point, inequality