The post How to Calculate Z-Score?: Statistics appeared first on MathSux^2.

]]>A normal curve is a bell shaped curve that shows the distribution of data evenly spread with respect to the mean. If you look at the normal curve below, the area under the curve shows all the possible probabilities of a certain data point occurring, notice the curve is higher towards the center mean, μ, and gets smaller as the distance from μ grows. The distance from μ is measured by the standard deviation, a unique unit of measurement that is specific to each group of data.

**Mean**: The mean always falls directly in the center of our normal curve. It is the average of our data, and always falls right in the middle.

**Standard Deviation**: This value is used as a standard unit of measurement for the data, measuring the distance between each data point in relation to the mean throughout the entire data set. For a review on what standard deviation is and how to calculate it, check out this post here.

Now for our normal curve:

Notice half of the data is below the mean, μ, while the other half is above? The normal curve is symmetrical about the mean, μ!

Z Score can tell us at what percentile a certain point in the data set falls in relation to the rest of the mean by using the standard deviation as a unit of measurement. If this sounds confusing, it’s ok! Take a look at the following formula:

We use the above formula in conjunction with a z table which tells us the probability under the curve for a certain point.

**Step 1: **First, let’s draw out our given information the mean=500, standard deviation=100, and the data point the question is asking for x=500 onto a normal curve. Notice that we want to find the value of the area under the curve shaded in pink. This will tell us the percent of students that scored below 500.

**Step 2: **We need to find the z-score by, using the data point given to us x=500, the mean=500, and the standard deviation, sigma=100.

**Step 3: **Yes, we have a zero! Now we need to take our **z table **and line up our chart. Notice that the chart finds the probability for everything at the beginning of the normal curve and on. This is perfect for answering our question!

**Step 4:** The table gives us our solution of .5000. If we multiply .5000 times 100 it gives us the percent of students who scored below 500 at 50%.

**Step 1: **First, let’s draw out our given information the mean=500, standard deviation=100, and the data point the question is asking for x=620 onto a normal curve. Notice that we want to find the value of the area under the curve shaded in pink. This will tell us the percent of students that scored above 620.

**Step 2: **We need to find the z-score by, using the data point given to us x=620, the mean=500, and the standard deviation, sigma=100.

**Step 3: **Yes, we got 1.2! Now we need to take our **z table **and line up our chart. Notice that the chart finds the probability for everything at the beginning of the normal curve and on. This is means to find the percent we are looking for, we need to subtract our answer from one since we want the value of probability on the right side of the curve (the z-table only provides the left side).

**Step 4: **The table gives us our solution of .8849. If we subtract this value from 1 then multiply that value times 100 it gives us the percent of students who scored above 620.

**Step 1:** In this question we have to work backwards by first identifying, where on the z-score table is the number .1611 and then filling in our z score formula to find x, the missing data point (in this case test score).

Search the table for .1611:

Notice that .1611 can be found on the z-table above with z-score -0.99. This is what we’ll use to find the unknown data point!

**Step 2: **We need to find the unknown test score by, using the z score we just found z=-0.99, the mean=500, and the standard deviation, sigma=100.

**Step 3: **Solve for x.

The grades on a final English exam are normally distributed with a mean of 75 and a standard deviation of 10.

a) What percent of students scored below a 60?

b) What percent of students scored above an 89?

c) What is the highest possible grade that included in the 4.46^{th} percentile?

d) What percent of students got at least a 77?

a) 6.68%

b) 8.08%

c) 58

d) 42.07%

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]]>The post Simultaneous Equations: Algebra appeared first on MathSux^2.

]]>**Simultaneous Equations** are when two equations are graphed on a coordinate plane and they intersect at, at least one point. The coordinate point of intersection for both equation is the answer we are trying to find when solving for simultaneous equations. There are three different methods for finding this answer:

We’re going to go over each method for solving simultaneous equations step by step with the example below:

The idea behind **Substitution**, is to solve for 1 variable first algebraically, and the plug this value back into the other equation solving for one variable. Then solving for the remaining variable. If this sounds confusing, don’t worry! We’re going to do this step by step:

**Step 1: **Let’s choose the first equation and move our terms around to solve for y.

**Step 3: **The equation is set up and ready to solve for x!

**Step 4: **All we need to do now, is plug x=3 into one of our original equations to solve for y.

**Step 5: **Now that we have solved for both x and y, we have officially found where these two simultaneous equations meet!

The main idea of **Elimination** is to add our two equations together to cancel out one of the variables, allowing us to solve for the remaining variable. We do this by lining up both equations one on top of the other and adding them together. If variables at first do not easily cancel out, we then multiply one of the equations by a number so it can. Check out how it’s done step by step below!

**Step 1: **First, let’s stack both equations one on top of the other to see if we can cancel anything out:

**Step 2: **Our goal is to get a 2 in front of y in the first equation, so we are going to multiply the entire first equation by 2.

**Step 3: **Now that we multiplied the entire first equation by 2, we can line up our two equations again, adding them together, this time canceling out the variable y to solve for x.

**Step 4: **Now, that we’ve found the value of variable x=3, we can plug this into one of our equations and solve for missing unknown variable y.

**Step 5: **Now that we have solved for both x and y, we have officially found where these two simultaneous equations meet!

The main idea of **Graphing **is to graph each a equation on a coordinate plane and then see at what point they intersect. This is the best method to visualize and check our answer!

**Step 1: **Before we start graphing let’s convert each equation into y=mx+b (equation of a line) form.

Equation 1:

Equation 2:

**Step 2: **Now, let’s graph each line, y=3x-4 and y=-x+8, to see at what coordinate point they intersect.

Need to review how to draw an equation of a line? Check out this post here! Notice we got the same exact answer using all three methods (1) Substitution (2) Elimination and (3) Graphing.

Ready to try the practice problems on your own?! Check them out below!

Solve the following simultaneous equations for x and y.

- (1, 3)
- (4,5)
- (-1, -6)
- (3, -3)

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]]>The post 30 60 90 Special Triangles: Geometry appeared first on MathSux^2.

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The 30 60 90 triangle is special because it forms an equilateral triangle when a mirror image of itself is drawn, meaning all sides are equal! This allows us to find the ratio between each side of the triangle by using the Pythagorean theorem. Check it out below!

Now let’s draw a mirror image of our triangle. Next, we can label the length of the new side opposite 30º “a,” and add this new mirror image length with the original we had to get, a+a=2a.

If we look at our original 30 60 90 triangle, we now have the following values for each side based on our equilateral triangle:

Now we can re-label our triangle, knowing the length of the hypotenuse in relation to the two legs. This creates a **ratio** that applies to all 30 60 90 triangles!

Knowing the above ratio, allows us to find any length of any and every 30 60 90 triangle, when given the value of one of its sides.

Let’s try an **Example**:

-> First let’s look at our ratio and compare it to our given triangle.

->Notice we are given the value of a, which equals 4, knowing this we can now fill in each length of our triangle based on the ratio of a 30 60 90 triangle.

Now let’s look at an **Example** where we are given the length of the hypotenuse and need to find the values of the other two missing sides.

->First let’s look at our ratio and compare it to our given triangle.

-> Notice we are given the value of the hypotenuse, *2a=20*. Knowing this we can find the value of *a* by dividing 20 by 2 to get *a=10*. Once we have the value of a=10, we can easily find the length of the last leg based on the 30 60 90 ratio:

Now for our last **Example**, when we are given the side length across from 60º and need to find the other two missing sides.

->First let’s look at our ratio and compare it to our given triangle.

-> In this case, we need to use little algebra to find the value of a, using the ratio for 30 60 90 triangles.

Now that we have one piece of the puzzle, the value of a, let’s fill it in our triangle below:

Finally, let’s find the value of the length of the hypotenuse, which is equal to 2a.

Find the value of the missing sides of each 30 60 90 degree triangle.

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above. Happy calculating!

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*Looking to review 45 45 90 degree special triangles? Check out this post here!*

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]]>The post NYE Ball Fun Facts: Volume & Combinations appeared first on MathSux^2.

]]>Yes, apparently the shape of the New Year’s ball is officially called a “Geodesic Sphere.” It is 12 feet in diameter and weighs 11,875 pounds.

If we wanted to **estimate** the volume of the New Year’s Ball we would could use the formula for volume of a sphere:

Talk about the ultimate shiny bauble! The NYE ball lights up the night with all 2,688 crystals in the shape of different sized triangles, each with heights of 5.75 inches or 4.75 inches.

On each triangle, there are 48 LEDs: 12 red, 12 blue, 12 green, and 12 white, for a total of 32,256 LEDs on the entire NYE ball itself.

**Permutations**: With this many lights and colors, there are **over a billion **potential permutations of colors on the entire NYE ball.

**Combinations**: Let’s break down one triangle with 48 LED lights each with 12 red, 12 blue, 12 green, and 12 white LEDs. How many possible combinations of lights are possible if we were to choose 7 blue, 5 red, 10 green, and 1 white turned on all at the exact same time?

We end up with the combination formula below:

That means that there are 496,793,088 possible ways that 7 blue lights, 5 red lights, 10 green lights, and 1 white light can be lit up on a triangle that is part of the entire NYE ball!

Interested in more NYE fun facts? Check out the sources of this article here.

NYE Fact Sheet from: timessquarenyc.org

NYE Ball picture: Timesquareball.net

If you like finding the volume of the NYE ball maybe, you’ll want to find the volume of the Hudson Yards Vessal in NYC here. Happy calculating and Happy New Year from MathSux!

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]]>The post Trig Functions (Amp, Freq, Phase Shifts): Algebra 2/Trig. appeared first on MathSux^2.

]]>*For a review on how to derive basic Trig functions (y=sinx, y=cosx, and y=tanx), click here.

**Amplitude**: The distance (or absolute value) between the *x*-axis and the highest point on the graph.

**Frequency: **This is the number of cycles that happen between 0 and 2π. (Α “cycle” in this case is the number of “s” cycles for the sine function).

**Period: **The x-value/length of one cycle. (Α “cycle” in this case is the number of “s” cycles for the sine function). This is found by looking at the graph and seeing where the first cycle ends, or, by using the formula:

**Horizontal Shift: **When a trigonometric function is moved either *left* or *right* along the x-axis.

**Vertical Shift:** When a trigonometric function is moved either *up* or *down* along the y-axis.

Let’s try an **Example**, graphing a Trig Function step by step.

**Step 1: **First let’s label and identify all the different parts of our trig function.

**Step 2: **Now let’s transform our graph one step at a time. First let’s start graphing y=cos(x) without any transformations.

**Step 3: **Let’s add our amplitude of 2, the distance to the x-axis. To do this our highest and lowest points on the y-axis will now be moved to 2 and -2 respectively.

**Step 4: **Next, we do a horizontal phase shift to the left by (π/2). To do this, we look at where negative (π/2) is on our graph at (-π/2) and move our entire graph over to start at this new point, “shifting” it to (π/2).

**Step 5: **For our last transformation, we have a vertical phase shift up 1 unit. All this means is that we are going to shift our entire graph up by 1 unit along the y-axis.

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above for an in depth explanation. Happy calculating!

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]]>The post Graphing Linear Inequalities: Algebra appeared first on MathSux^2.

]]>When graphing linear inequalities, we always want to treat the inequality as an equation of a line in form y=mx+b….with a few exceptions:

Now that we know the rules, of graphing inequalities, let’s take a look at an **Example**!

**Step 1:** First, let’s identify what type of inequality we have here. Since we are working with a > sign, we will need to use ** a dotted line **and

**Step 2: **Now we are going to start graphing our linear inequality as a normal equation of a line, by identifying the slope and the y-intercept only this time keeping open circles in mind. (For a review on how to graph regular equation of a line in y=mx+b form, click here)

**Step 3: **Now let’s connect our dots, by using a dotted line to represent our greater than sign.

**Step 4: **Now it is time for us to shade our graph, since this is an inequality, we need to show all of our potential solutions with shading. Since we have a greater than sign, , we will be shading above the y-axis. Notice all the positive y-values above are included to the left of our line. This is where we will shade.

**Step 5: **Check! Now we need to check our work. To do that, we can choose any point within our shaded region, if the coordinate point we chose hold true when plugged into our inequality then we are correct!

Let’s take the point (-3,2) plugging it into our inequality where x=-3 and y=2.

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above for an in depth explanation. Happy calculating!

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]]>The post Central Angles Theorems: Geometry appeared first on MathSux^2.

]]>Central angles and arcs form when two radii are drawn from the center point of a circle. When these two radii come together they form a **central angle**. A central angle is equal to the length of the arc. When it comes to measuring the central angle, the central angle is always equal to arc length and vice versa:

Central Angles = Arc Length

There are a two central angle theorems to know, check them out below!

**Central Angle Theorem #1:**

**Central Angle Theorem #2:**

Let’s look at how to apply these rules with an **Example:**

Let’s do this one step at a time.

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above. Happy calculating!

*Also, if you want to check out **Intersecting Secants** click this link here!

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]]>The post TikTok Math Video Compilations appeared first on MathSux^2.

]]>Want to make math suck just a little bit less? Subscribe and follow us for FREE fun colorful math videos and lessons every week!

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Within algebra, you will find arithmetic sequences, combining like terms, box and whisker plots, geometric sequences, solving radical equations, completing the square, 4 ways to factor quadratic equations, piecewise functions and more!

Within Geometry, you will find, how to construct an equilateral triangle, a median of a trapezoid, area of a sector, how to find perpendicular and parallel lines through a given point, SOH CAH TOA right triangle trigonometry, reflections, and more!

Within Algebra 2/Trig., you will find, how to expand a cubed binomial, how to divide polynomials, how to solve log equations, imaginary numbers, synthetic division, unit circle basics, how to graph y=sin(x), and more!

Within statistics, you will find, box and whisker plots, how to find the variance, and, the probability of flipping a coin 2 times!

For full length video, don’t forget to check out our free math video index page! Thanks for stopping by!

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]]>The post Graphing Trig Functions: Algebra 2/Trig. appeared first on MathSux^2.

]]>For deriving our trigonometric function graphs [y=sin(x), y=cos(x), and y=tan(x)] we are going to write out our handy dandy **Unit Circle.** By looking at our unit circle and remembering that coordinate points are in (cos(x), sin(x)) form and that tanx=(sin(x))/(cos(x)) we will be able to derive each and every trig graph!

*Note below is the unit circle we are going to reference to find each value, for an in depth explanation of the unit circle, check out this link here.

**Step 1: **We are going to derive each degree value for sin by looking at the unit circle. These will be our coordinates for graphing y=sin(x). *For a review on how to get these values, check out the link here explaining the unit circle.

**Step 2: **Now we need to convert all the from degrees to radians. Fear not because this can be done easily with a simple formula!

** **To convert degrees à radians, just use the formula below:

**Step 3: **Now that we have our coordinate points and converted degrees to radians, we can draw out our function y=sin(x) on the coordinate plane!

Now we will follow the same process for graphing y=cos(x) and y=tan(x).

**Step 1: **We are going to derive each degree value for cos by looking at the unit circle. These will be our coordinates for graphing y=cos(x). *For a review on how to get these values, check out the link here explaining the unit circle.

**Step 2:** Now we need to convert all the from degrees to radians.

**Step 3: **Now that we have our coordinate points and converted degrees to radians, we can draw out our function y=cos(x) on the coordinate plane!

**Step 1: **We are going to derive each degree value for tan by looking at the unit circle. In order to derive values for tan(x), we need to remember that tan(x)=sin(x)/cos(x). Once found, these will be our coordinates for graphing y=tan(x). *For a review on how to get these values, check out the link here explaining the unit circle.

**Step 2:** Now we need to convert all the from degrees to radians.

**Step 3: **Now that we have our coordinate points and converted degrees to radians, we can draw out our function y=tan(x) on the coordinate plane!

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above for an in depth explanation. Happy calculating!

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]]>The post Geometric Sequences: Algebra appeared first on MathSux^2.

]]>**Geometric sequences **are a sequence of numbers that form a pattern when the same number is either *multiplied* or *divided* to each subsequent term.

Example:

Notice we are multiplying 2 to each term in the sequence above. If the pattern were to continue, the next term of the sequence above would be 64. This is a geometric sequence!

In this sequence it’s easy to see what the next term is, but what if we wanted to know the 15^{th} term? That’s where the Geometric Sequence formula comes in!

Now that we broke down our geometric sequence formula, let’s try to answer our original question below:

->First, let’s write out the formula:

-> Now let’s fill in our formula and solve with the given values.

Let’s look at another example where, the common ratio is a bit different, and we are dividing the same number from each subsequent term:

-> First let’s identify the common ratio between each number in the sequence. Notice each term in the sequence is being divided by 2 (or multiplied by 1/2 ).

-> Now let’s write out our formula:

-> Next let’s fill in our formula and solve with the given values.

- Find the 12
^{th}term given the following sequence: 1250, 625, 312.5, 156.25, 78.125, …. - Find the 17
^{th}term given the following sequence: 3, 9, 27, 81, 243,….. - Find the 10
^{th}term given the following sequence: 5000, 1250, 312.5, 78.125 ….. - Shirley has $100 that she deposits in the bank. She continues to deposit twice the amount of money every month. How much money will she deposit in the twelfth month at the end of the year?

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above. Happy calculating!

*Also, if you want to check out **arithmetic sequences** click this link here!

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]]>