Because of natural variability in manufacturing, a 16-ounce container of oats does not usually hold exactly 16 ounces of oats. A container is permitted to hold a little more or a little less. The specifications for the oat-filling machine are that it needs to fill each container with 16 ounces of oats with a 2% margin of error. If a container is filled with 16.5 ounces of oats, is the filling machine working within specifications? Explain.

Increasing [tex]\frac{15}{16}[/tex] by [tex]\frac{3}{4}[/tex] of it and then increasing the result by [tex]\frac{3}{5}[/tex] of it gives us the fraction of [tex]\frac{45}{32}[/tex].

To increase [tex]\frac{15}{16}[/tex] by [tex]\frac{3}{4}[/tex] of it, we can multiply [tex]\frac{15}{16}[/tex] by [tex]\frac{3}{4}[/tex], which gives us:

[tex]\frac{15}{16}[/tex] × [tex]\frac{3}{4}[/tex] = [tex]\frac{45}{64}[/tex]

Adding this result to [tex]\frac{15}{16}[/tex], we get:

[tex]\frac{15}{16}[/tex] + [tex]\frac{45}{64}[/tex] = [tex]\frac{225}{256}[/tex]

Now, to increase this result by [tex]\frac{3}{5}[/tex] of it, we can multiply [tex]\frac{225}{256}[/tex] by 3/5, which gives us:

[tex]\frac{225}{256}[/tex] × [tex]\frac{3}{5}[/tex] = [tex]\frac{135}{256}[/tex]

Adding this result to [tex]\frac{225}{256}[/tex], we get the final answer:

[tex]\frac{225}{256}[/tex] + [tex]\frac{135}{256}[/tex] = [tex]\frac{360}{256}[/tex]

Simplifying this fraction, we get:

[tex]\frac{360}{256}[/tex] = [tex]\frac{45}{32}[/tex]

The problem involves increasing a fraction by a certain percentage twice. To solve the problem, we first find the increase in the fraction by multiplying it with the given percentage. We then add the result to the original fraction to get the new fraction. This process is repeated for the second increase. Finally, we simplify the resulting fraction if possible. In this specific problem, we are increasing [tex]\frac{15}{16}[/tex] by [tex]\frac{3}{4}[/tex] of it and then increasing the result by [tex]\frac{3}{5}[/tex] of it to find the resulting fraction [tex]\frac{45}{32}[/tex].

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Answer:

11/3

Step-by-step explanation:

2[tex]\frac{1}{5}[/tex] = 11/5

2[tex]\frac{1}{5}[/tex] ÷ [tex]\frac{3}{5}[/tex] = 11/5 · 5/3 = 55/15 = 11/3

Answer:

[tex]\large\boxed{(2, -3)}[/tex]

Step-by-step explanation:

[tex]\textsf{We are asked for the Center Coordinates given a circle.}[/tex]

[tex]\large\underline{\textsf{What is a Circle?}}[/tex]

[tex]\textsf{A Circle is a curved shape that is commonly used to present a equidistant (same}[/tex]

[tex]\textsf{distance apart) from a Center point.}[/tex]

[tex]\textsf{When we create an equation for a circle, we should keep in mind of what the}[/tex]

[tex]\textsf{coordinates of the center point are, and the Radius. For this problem an equation}[/tex]

[tex]\textsf{has been given to us.}[/tex]

[tex]\boxed{\begin{minipage}{20 em} \\ \underline{\textsf{\large Equation of a Circle (Standard Form);}} \\ \\ \tt \tt (x-h)^{2} + (\tt y-k)^{2} = r^{2} \\ \\ \underline{\textsf{\large Where;}} \\ \textsf{h represents the x-coordinate of the center.} \\ \textsf{k represents the y-coordinate of the center.} \\ \tt r^{2} \ \textsf{represents the radius of a circle squared.} \end{minipage}}[/tex]

[tex]\textsf{*Note that if we are given the formula, the radius is already squared, unless}[/tex]

[tex]\textsf{specified in the directions.}[/tex]

[tex]\tt (x-2)^{2} + (y+3)^{2} = 9[/tex]

[tex]\textsf{Our main focus should be h and k. Notice how both point values are negative.}[/tex]

[tex]\textsf{The Center Point is represented as the opposite given to us in the formula.}[/tex]

[tex]\underline{\textsf{Find the Opposite of -2 and 3;}}[/tex]

[tex]\tt -2 \rightarrow \boxed{2}[/tex]

[tex]3 \rightarrow \boxed{-3}[/tex]

[tex]\underline{\textsf{Center Coordinates;}}[/tex]

[tex]\large\boxed{(2, -3)}[/tex]

Answer:

b

Step-by-step explanation:

just add then then after you add 27=27 multiply it by 2

Answer:

let n be the number

6 less than the number

= (n-6)^5 = 32

(n-6)^5 = 2^5

since powers are equal we equate the bases

n-6 =2

n=8

Answer:

36.4

Step-by-step explanation:

In morning, builder worked for = 25 in a hour.

Let hour be ? hr.

then 25? = Time he worked in morning.

In afternoon,he worked for= 910 of ? hour = 910? hr

Times he worked more = 910?/25? = 36.4 times

25*6 =

Answer:

A random number list could be used to simulate the probability of rain each day this week. For example, we could assign the numbers 1-60 to the days where rain is expected and the numbers 61-100 to the days where it is not expected. Then, we could use a random number generator to simulate each day and count how many days it takes until it rains for the first time. By repeating this simulation many times and taking the average, we could estimate the experimental probability of it taking at least three days before it rains.

0 

a) The probability that the battery lasts more than 4 hours is 0.758 (rounded to 3 decimal places).

b) The 25% value is 240 minutes, and the 75% value is 310 minutes.

c) The value of life in minutes that is exceeded with 95% probability is 357 minutes.

a) To find the probability that the battery lasts more than 4 hours (240 minutes), we first need to convert 240 minutes into a z-score.
z = (X - μ) / σ
z = (240 - 275) / 50
z = -0.7
Now, we'll use a z-table to find the probability that the battery lasts more than 4 hours:
P(Z > -0.7) = 1 - P(Z ≤ -0.7) = 1 - 0.2420 = 0.758
The probability that the battery lasts more than 4 hours is 0.758 (rounded to 3 decimal places).
b) To find the quartiles, we'll use the z-table to find the z-scores corresponding to 25% and 75%:
25%: z = -0.674
75%: z = 0.674
Now, we'll convert the z-scores back into minutes:
Q1 = μ + z * σ = 275 + (-0.674) * 50 = 240 (rounded to the nearest integer)
Q3 = μ + z * σ = 275 + (0.674) * 50 = 310 (rounded to the nearest integer)
The 25% value is 240 minutes, and the 75% value is 310 minutes.
c) To find the value of life in minutes that is exceeded with 95% probability, we first find the z-score corresponding to 95%:
95%: z = 1.645
Now, we'll convert the z-score back into minutes:
X = μ + z * σ = 275 + (1.645) * 50 = 357 (rounded to the nearest integer).

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Hence, point P's coordinates are (6, 3) as the midpoint of the line segment from R to Q, which is the precise halfway point between R and Q.

A point's location on a line or in space can be determined using a set of numbers called coordinates. An x-coordinate and a como, which denote the horizontal as well as vertical ranges from a bench mark known as the origin, are the two numbers that make up coordinates in two-dimensional space. Three numbers, the x, y, and z coordinates, which describe the distances from origin in the x, y, and g directions, respectively, might make up coordinates in three-dimensional space. Among other disciplines, coordinates are frequently employed in physics, arithmetic, and geometry.

given

The midpoint of the line segment from R to Q, which is the precise halfway point between R and Q, must be located in order to determine the coordinates of point P.

We can separately aver the x- and y-coordinates to get the middle.

The midpoint's x-coordinate is (4 + 8)/2, which equals 6.

(0 + 6)/2 = 3 is the midpoint's y-coordinate.

Hence, point P's coordinates are (6, 3) as the midpoint of the line segment from R to Q, which is the precise halfway point between R and Q.

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Answer:

(-3.5 , 2.3)

Step-by-step explanation:

I know

Answer:

Step-by-step explanation:

14.516129%

Answer:

Step-by-step explanation:

Search up the area of both. rectangle is LxW therefore the area is 18.  the other is trapezium therefore, ((a+b)xh)/2 therefore the area is 20. trapezium has larger area

Answer trapezoid has  greater area

Step-by-step explanation: rec: 6*3= 8,  trapezoid [ (6+2) * 5 ]1/2  = 20

The proportion of Year 9 pupils in the music club is approximately 46.7%.

A ratio or value that may be stated as a fraction of 100 is called a percentage.

To find the proportion of Year 9 pupils in the music club as a percentage, follow these steps:

To get proportion as a percentage divied number of Year 9 pupils by total pupils multiplied by 100.

Proportion as a percentage = [tex]\frac{Year \:9 \:pupils}{total \:pupils} \times 100[/tex]
Divide the number of Year 9 pupils by the total number of pupils.

78 (Year 9 pupils) ÷ 167 (total pupils)

= 0.4671
Multiply the result by 100 to convert the proportion to a percentage.
0.4671 × 100

= 46.71%.

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Answer:

The equation of the given circle is (x-2)^2 + (y+3)^2 = 9, which is in the standard form of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.

Comparing the given equation with the standard form, we can see that the center of the circle is (2, -3) and the radius is 3.

Therefore, the radius of the circle is 3 units.

Step-by-step explanation:

The area of a right triangle like this is   area = 1/2 * base * height

   base and height are the LEGS of the right triangle

  this one pictured piece area is:        1/2 * 14 in  * 11 in = 77 in^2

    He needs 12 of them ....so total is then   12 * 77 in^2 = 924 in^2 needed

Answer:

f(-3/2) = -411/16.

Step-by-step explanation:

42/100 = 21/50 about 42%

The answer is y = ±1, as well as two complex solutions. The answer y = 3 is not a solution to the equation [tex]y^3 = 1.[/tex]

An equation is a statement that asserts the equality of two expressions. It typically contains one or more variables, which are symbols that represent unknown or varying quantities. The expressions on either side of the equals sign can include numbers, constants, functions, and other mathematical operations.

To solve[tex]y^3 = 1[/tex], we can take the cube root of both sides of the equation:

y = ∛1

There are three cube roots of 1, which are 1, -1/2 + √3/2 i, and -1/2 - √3/2 i, where i is the imaginary unit. These three roots form a complex conjugate pair.

Therefore, the solutions to the equation [tex]y^3 = 1[/tex] are:

y = 1

y = -1/2 + √3/2 i

y = -1/2 - √3/2 i

So the answer is y = ±1, as well as two complex solutions. The answer y = 3 is not a solution to the equation [tex]y^3 = 1.[/tex]

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Therefore, the correct option is C). graph with the x axis labeled number of days and the y axis labeled total amount of

sunglasses and a line going from the point 0 comma 52 through the point (1,49).

mathematics, a graph is a data structure that represents a collection of vertices (also known as nodes) and edges between them.

A graph can be used to represent a wide range of real-world systems, such as social networks, transportation networks, and electrical circuits.

In a graph, vertices are usually represented as points or circles, and edges are represented as lines or arrows that connect pairs of vertices. Each edge can be directed (pointing from one vertex to another) or undirected (connecting two vertices without any specific direction).

Graphs can be classified in many ways, such as by their density (the ratio of edges to vertices), their connectivity (how many separate components the graph has), and their degree distribution (the distribution of the number of edges connected to each vertex). Graph theory, which is the study of graphs, has many practical applications in computer science, mathematics, and other fields.

Which graph shows the relationship given in the table?

A) graph with the x axis labeled number of days and the y axis labeled total amount of

sunglasses and a line going from the point 0 comma 52 through the point 1 comma 50.

B) graph with the x axis labeled number of days and the y axis labeled total amount of

sunglasses and a line going from the point 0 comma 56 through the point (1,54)

C) graph with the x axis labeled number of days and the y axis labeled total amount of

sunglasses and a line going from the point 0 comma 52 through the point (1,49)

D) graph with the x axis labeled number of days and the y axis labeled total amount

of sunglasses and a line going from the point 0 comma 56 through the point (1,53)

Based on the given data in the table, the inventory of sunglasses decreases as the number of days since the opening of the eyeglasses store increases. This suggests a negative correlation between the two variables.

To visualize this relationship, we can plot the data points on a scatter plot and draw a line of best fit. Based on the options provided, the closest graph to the relationship given in the table is option C) with the x-axis labeled as number of days, the y-axis labeled as the total amount of sunglasses, and a line going from the point 0,52 through the point 1,49.

Option A) has a decreasing trend, but the line does not pass through all the given data points.

Option B) has a similar decreasing trend, but the values on the y-axis are higher than those in the table.

Option D) has a decreasing trend, but the values on the y-axis are higher than those in the table, and the line does not pass through all the given data points.

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A: 75% of items are classified as C items, 18.75% as B items, and 6.25% as A items.

A-B-C examination is an instrument utilized for stock administration, which orders things into three classifications: A, B, and C, in light of their worth and significance. Classification A things have a high worth however a low volume, and their administration requires close consideration and control. Class B things have moderate worth and volume and require moderate control. Classification C things have a low worth yet a high volume, and their administration can be loose. The reason for the examination is to recognize which things need the most consideration and control, and which things can be dealt with less consideration and cost-actually. The examination assists in streamlining with reviewing levels, decreasing expenses, and expanding benefits.

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The selling price of the article without VAT is Rs. 6,429.13.

What is Algebraic expression ?

In mathematics, an algebraic expression is a combination of variables, constants, and mathematical operations, such as addition, subtraction, multiplication, and division, that represents a quantity or a relationship between quantities.

Let's assume the rate of discount to be 'x' and the rate of VAT to be 'y'.

According to the problem, the marked price of the article is Rs. 8,000.

After applying the discount, the selling price becomes:

Selling price = Marked price - Discount

Selling price = 8000 - (x÷100) * 8000

Now, we add the VAT to the selling price:

Selling price with VAT = Selling price + (y÷100) * Selling price

Selling price with VAT = [8000 - (x÷100) * 8000] + [(y÷100) * (8000 - (x÷100) * 8000)]

Selling price with VAT = 6689.60 (Given)

We are also given that the rate of discount was double the rate of VAT, i.e.,

x = 2y

Substituting this value in the above equation, we get:

6689.60 = [8000 - (2y÷100) * 8000] + [(y÷100) * (8000 - (2y÷100) * 8000)]

Simplifying this equation, we get:

6689.60 = 8000 * [1 - (2y÷100) + (y÷100) - (2y÷100) * (y÷100)]

6689.60 = 8000 * [1 - (3y÷50) + (y*y÷5000)]

Solving for y, we get:

y = 4

Now, we can calculate the selling price without VAT as follows:

Selling price without VAT = Selling price - (y÷100) * Selling price

Selling price without VAT = 6689.60 - (4÷100) * 6689.60

Selling price without VAT = Rs. 6,429.13

Therefore, the selling price of the article without VAT is Rs. 6,429.13.

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Answer:

Option D is the correct answer

D. f(n) = 95 + 60(n - 1)

Step-by-step explanation:

Solution:

A number line that represent the values of x is shown in the graph attached below.

In Mathematics and Geometry, a number line simply refers to a type of graph with a graduated straight line which comprises both positive and negative numbers that are placed at equal intervals along its length.

This ultimately implies that, a number line primarily increases in numerical value towards the right from zero (0) and decreases in numerical value towards the left from zero (0):

x < - 1 or x > 1

-1 > x > 1

-3 < x ≤ 1

-5 ≤ x ≤ 0

In this scenario and exercise, we would use an online graphing calculator to plot each of the inequality as shown in the graph (number line) attached below.

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The corresponding inequality for this graph is

0 <= x <= 5

An inequality is a statement that two values are not necessarily equal. In other words, it expresses a relation between two quantities that can be either less than, greater than, less than or equal to, or greater than or equal to each other. Inequalities are typically written using symbols such as "<" (less than), ">" (greater than), "≤" (less than or equal to), and "≥" (greater than or equal to).

The graph represents a shaded rectangle with sides on the x and y axes, and with one corner at the origin (0,0). The top-right corner of the rectangle is at the point (5,5), and the sides of the rectangle are parallel to the axes.

The corresponding inequality for this graph is 0 <= x <= 5, which can be read as "x is greater than or equal to 0 and less than or equal to 5".

Therefore, the correct answer is:

0 <= x <= 5

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The complete question is:

Which of the following inequalities matches the graph?

Ox≤5

0<=x<=5

The value of x is 29.5 units. Since the segments inside the circle have the same length, we can set  52.5  - x = 23 and get the value of x.

It is half of the diameter of the circle. The radius is used to calculate the area of a circle and the length of an arc or a circular sector.

In order to find the value of x, we need to first determine the radius of the circle.

In this problem, the first shorter side is 45 units, and the second shorter side is 27 units. Thus, we can find the radius of the circle by using the following method,

Radius of the circle = √(45² + 27²)

= √(2754) = 52.5 units

Now that we know the radius of the circle, we can use this information to find the value of x. We know that the total length of the first segment is  52.5 units. This means that the length of the segment outside the circle is x units and the length of the segment inside the circle is 52.5  - x units.

Similarly, the total length of the second segment is 50 units, so the length of the segment outside the circle is 27 units and the length of the segment inside the circle is

50 - 27 = 23 units.

Since the segments inside the circle have the same length, we can set  52.5  - x = 23 and solve for x.

52.5  - x = 23

x =  52.5  - 23

x = 29.5 units

Therefore, the value of x is 29.5 units.

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Answer:

(2,1)          (2,1)

1=5(2)-9   1=-2(2)+5

1=10-9      1=-4+5

Step-by-step explanation:

Hope this helps!!

Mark me brianliest :))

1. The cannonball is at 70 meters before it is launched.

2. The cannonball reaches its maximum height at about 3 seconds.

3. The maximum height of the cannonball is about 115 meters.

4. It takes about 8 seconds for the cannonball to reach the ground.

I don't know the answer to number 5 - sorry :(

I hope the rest are correct

Answer:

A

Step-by-step explanation:

Given:

r (radius) = 5 in

h (height) = 5 in

Find: A (surface area) - ?

.

[tex]a(surface) = a(lateral) + a(2 \: bases)[/tex]

First, we have to find the lateral surface:

[tex]a(lateral) = 2\pi \times r \times h = 2 \times 3.14 \times 5 \times 5 = 157 \: {in}^{2} [/tex]

Then, we have to find the area of the base and multiply it by 2, since the shown figure is a cylinder, which has 2 identical bases:

[tex]a(2 \: bases) = 2 \times \pi {r}^{2} = 2 \times 3.14 \times {5}^{2} = 157 \: {in}^{2} [/tex]

Now, let's add everything together and we'll get the answer:

[tex]a(surface) = 157 + 157 = 314 \: {in}^{2} [/tex]

x² + y² + 4x + 6y - 3 = 0  equation represents the general form a circle with a center at (â€"2, â€"3) and a diameter of 8 units.

The equation of a circle with center (h, k) and radius r can be written in the form (x - h)² + (y - k)² = r².

Given a center of (-2, -3) and a diameter of 8 units, first find the radius, which is half of the diameter:
r = 8 / 2 = 4
Now, substitute the center coordinates and radius into the equation:
(x - (-2))² + (y - (-3))² = 4²
(x + 2)² + (y + 3)² = 16
Now, expand the equation to get the general form:
(x² + 4x + 4) + (y² + 6y + 9) = 16
x² + 4x + y² + 6y + 4 + 9 - 16 = 0
Combine like terms to get the final equation:
x² + y² + 4x + 6y - 3 = 0

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