I need some help please
Given:-
[tex] \texttt{c = 1}[/tex][tex] \: [/tex]
[tex] \texttt{d = 3}[/tex][tex] \: [/tex]
Solution:-
[tex] \texttt{3(2c + d ) -4 ( c - d ) + d² }[/tex][tex] \: [/tex]
put the given values in tha equation
[tex] \texttt{= 3( 2 ( 1 ) + 3 ) -4 ( 1 - 3 ) + ( 3 )² }[/tex][tex] \: [/tex]
[tex] \texttt{= 3 ( 2 + 3 ) -4 ( -2 ) + 9}[/tex][tex] \: [/tex]
[tex] \texttt{= 3 ( 5 ) -4 ( -2 ) + 9}[/tex][tex] \: [/tex]
[tex] \texttt{= 15 + 8 + 9}[/tex][tex] \: [/tex]
[tex] \texttt{= 23 + 9}[/tex][tex] \: [/tex]
[tex] \boxed{ \texttt{ \purple{= 32}}}[/tex][tex] \: [/tex]
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Please help, just need H!
Answer:
Step-by-step explanation:
[tex]f(x)=\frac{27}{4} x[/tex]
[tex]p(x)=\sqrt{3x}[/tex]
[tex][p(x)]^6=(3^\frac{1}{2}x^\frac{1}{2})^6= 27x^3[/tex]
[tex][p(x)]^6g(x)=27x^3(\frac{x}{4} )=\frac{27x^4}{4}[/tex]
[tex][p(x)]^6g(x)[t(x)]^3=\frac{27x^4}{4} (\frac{1}{x} )^3=\frac{27x^4}{4x^3} =\frac{27}{4} x[/tex]
Solution: [tex][p(x)]^6g(x)[t(x)]^3=f(x)[/tex]
I think there might be nicer solutions but this works!
What method could have been used to collect the data
The method used to collect data will depend on the specific research question being asked and the resources available to the researcher.
One common method for collecting data is through experimentation. This involves designing a controlled experiment that allows researchers to manipulate one or more variables while keeping all other factors constant.
Observational data can be particularly useful when it is not possible or ethical to manipulate variables in an experiment.
Surveys and questionnaires are another way to collect data. These methods involve asking people to provide information about their attitudes, beliefs, and behaviors.
Other methods for collecting data include case studies, in which researchers closely examine a single individual or group, and archival research, in which data is collected from existing records, such as census data or medical records.
By carefully selecting the right method for their research, scientists can gather accurate and reliable data that can help to advance our understanding of the world around us.
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A regular hexagon has side lengths of 2x and has a perimeter that is equal to an isosceles triangle with legs of x + 4 and a base that is 3 less than one of the legs. Write and solve equation to find the length of one side of the hexagon.
According to the formula, the length of one side of the hexagon is 21/4 or 5.25 units.
What is the perimeter of the regular hexagon?
Let's call the length of one side of the regular hexagon "s". Since a hexagon has six sides, the perimeter of the hexagon is 6s.
According to the problem, this is equal to the perimeter of an isosceles triangle with legs of x+4 and a base that is 3 less than one of the legs.
The perimeter of the isosceles triangle is given by:
perimeter = 2(x+4) + (x+1)
where (x+1) is the length of the base (which is 3 less than one of the legs).
Setting the perimeters equal, we get:
6s = 2(x+4) + (x+1)
6s = 3x + 9
s = (3/2)x + (3/4)
We also know that the side lengths of the hexagon are 2x. Substituting this into the equation for "s", we get:
2x = (3/2)x + (3/4)
x = 3
Therefore, the length of one side of the hexagon is:
s = (3/2)x + (3/4) = (3/2)(3) + (3/4) = 9/2 + 3/4 = 21/4
So the length of one side of the hexagon is 21/4 or 5.25 units.
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The graph below shows petrol prices at two petrol stations, Station X and Station
Y.
Ellie went to one of the petrol stations and bought 20 litres of petrol for £24.
a) Did Ellie go to Station X or Station Y?
b) How much would 15 litres of petrol cost at the same station?
Give your answer in pounds (£).
Cost (£)
Cost against amount of petrol
40
30
20
10-
5
10 15 20 25
Amount of petrol (litres)
Key
Station X
Station Y
At Station Y, 15 litres of gasoline would cost £18.
What is cost in?Cost is the amount of money spent by a business to produce or create goods or services. It excludes the profit margin markup. Cost is the sum of money spent on making a good or product, as seen from the seller's perspective.
Ellie must have visited Station Y because she paid $24 for 20 litres of gasoline, proving that she did.
b) We can observe from the graph that 20 litres of gasoline at Station Y costs £24. With the help of this data, we can calculate how much a litre of gasoline costs:
Cost of 1 litre of petrol = Cost of 20 litres of petrol / 20
Cost of 1 litre of petrol = £24 / 20
Cost of 1 litre of petrol = £1.20
Therefore, 15 litres of petrol at Station Y would cost:
Cost of 15 litres of petrol = Cost of 1 litre of petrol x 15
Cost of 15 litres of petrol = £1.20 x 15
Cost of 15 litres of petrol = £18
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On January 1,1999 , the average price of gasoline was $1.19 per gallon. If the price of gasoline increased by 0.3% per month, which equation models the future cost of gasoline? y=1.19(1.003)^(x) y=1.19(x)^(1.03) y=1.19(1.03)^(x)
Answer:
first one
Step-by-step explanation:
The equation that models the future cost of gasoline is y=1.19(1.003)^(x), where "y" represents the future cost of gasoline per gallon and "x" represents the number of months since January 1, 1999.
In this equation, the initial cost of gasoline is $1.19 per gallon, and the cost increases by 0.3% per month, which is represented by the factor of (1.003)^(x).
Using this equation, you can calculate the future cost of gasoline for any number of months after January 1, 1999. For example, if you want to calculate the cost of gasoline 24 months after January 1, 1999, you can plug in x=24 and calculate y as follows:
y = 1.19(1.003)^(24)
y = 1.19(1.08357)
y = 1.288 per gallon
Therefore, the predicted cost of gasoline 24 months after January 1, 1999 is $1.288 per gallon.
HELP PLS combine the like terms 3x+5-x+3+4x
Answer:
3x, 4x | 5, 3
Step-by-step explanation:
If P = 2y² + 4xy + 4
Q = − 3y² + 7 - 3xy
R=- 3xy + 8
Find P+Q=R.
Answer:
P = [tex]2y^{2}[/tex] + 4xy +4
Q = [tex]-3y^{2}[/tex] + 7 -3xy
R = -3xy +8
Step-by-step explanation:
Anyone know the answer?
As a result, the Styrofoam collar has a volume of roughly 179.594 cubic inches.
what is volume ?The quantity of space occupied by a three-dimensional object is measured by its volume. Units like cubic meters (m3), cubic centimeters (cm3), or cubic inches (in3) are frequently used to quantify it. Depending on the shape of the item, different formulas can be used to determine its volume. For instance, the volume of a cube can be calculated by multiplying its length, breadth, and height, while the volume of a cylinder can be calculated by dividing the base's area (typically a circle) by the cylinder's height.
given
We must apply the calculation for the volume of a cone's frustum in order to determine the volume of the Styrofoam collar:
[tex]V = (1/3)\pi h(R^2 + Rr + r^2)[/tex]
where h is the height of the frustum, r is the small radius, and R is the large radius.
Given the numbers, we can determine:
R = 5 in.
3 centimeters is r.
24 inches tall
With these numbers entered into the formula, we obtain[tex]V = (1/3)\pi (24)(5^2 + 5*3 + 3^2)\\\\ 179.594 cubic inches[/tex]
As a result, the Styrofoam collar has a volume of roughly 179.594 cubic inches.
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Five percent of the parts produced by a machine are defective. Fifteen parts are selected at random. Use the binomial probability tables to answer the following questions. (a) What is the probability that exactly 3 parts will be defective? (Round your answer to four decimal places.) (b) What is the probability that the number of defective parts will be more than 2 but fewer than 6? (Round your answer to four decimal places.) (c) What is the probability that fewer than 3 parts will be defective? (Round your answer to four decimal places.) (d) What is the expected number of defective parts? (e) What is the variance for the number of defective parts?
The variance for the number of defective parts is 0.7125.
(a) What is the probability that exactly 3 parts will be defective? (Round your answer to four decimal places.)In the given question, the probability of a part being defective is 5%, which is represented as p=0.05. The probability of a part being non-defective is (1 - 0.05) = 0.95.The given question represents a binomial experiment, which includes the following conditions:The experiment consists of n identical trials.Each trial has only two possible outcomes, a success, and a failure.Success has probability p and failure has probability 1-p.The trials are independent.To calculate the probability of exactly 3 parts being defective when 15 parts are selected randomly, we use the following formula:P (X = k) = nCk x pk x (1-p) n-kWhere P(X=k) is the probability of getting k defective parts, nCk is the number of ways of getting k defects from n parts, pk is the probability of getting k defective parts, and (1-p) n-k is the probability of getting n-k non-defective parts.p = 0.05q = (1 - 0.05) = 0.95n = 15a. P (X = 3) = 15C3 x 0.05³ x 0.95¹² = 0.2508The probability of getting exactly 3 defective parts is 0.2508. Hence, the required probability is 0.2508.(b) What is the probability that the number of defective parts will be more than 2 but fewer than 6? (Round your answer to four decimal places.)b. We need to calculate the probability of getting defective parts more than 2 but less than 6.P (3 < X < 6) = P (X = 3) + P (X = 4) + P (X = 5)P (X = 3) = 15C3 x 0.05³ x 0.95¹² = 0.2508P (X = 4) = 15C4 x 0.05⁴ x 0.95¹¹ = 0.0925P (X = 5) = 15C5 x 0.05⁵ x 0.95¹⁰ = 0.0204P (3 < X < 6) = 0.2508 + 0.0925 + 0.0204 = 0.3637The probability of getting defective parts between 2 and 6 is 0.3637. Hence, the required probability is 0.3637.(c) What is the probability that fewer than 3 parts will be defective? (Round your answer to four decimal places.)c. We need to calculate the probability of getting fewer than 3 parts defective. The probability of getting zero defective parts or getting one defective part is given by:P (X = 0) = 15C0 x 0.05⁰ x 0.95¹⁵ = 0.4630P (X = 1) = 15C1 x 0.05¹ x 0.95¹⁴ = 0.3456P (X < 3) = P (X = 0) + P (X = 1) = 0.4630 + 0.3456 = 0.8086The probability of getting fewer than 3 defective parts is 0.8086. Hence, the required probability is 0.8086.(d) What is the expected number of defective parts?The expected number of defective parts is given by:μ = npμ = 15 × 0.05μ = 0.75The expected number of defective parts is 0.75.(e) What is the variance for the number of defective parts?The variance for the number of defective parts is given by:σ² = npqσ² = 15 × 0.05 × 0.95σ² = 0.7125The variance for the number of defective parts is 0.7125.
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Someone please help with this? Thank you!
Table values are -3, -1, 3, 5, 13
Define the term function?A function is a mathematical object that maps each element from one set to a unique element in another set. Functions are represented using symbols and can be described using graphs, tables, or equations.
Given function is,
[tex]f(x)=2x +3[/tex]
Solve for x = -3, f(-3) = 2×(-3) + 3 = -6 + 3 = -3
f(-3) = -3
Solve for x = -2, f(-2) = 2×(-2) + 3 = -4 + 3 = -1
f(-2) = -1
Solve for x = 0, f(0) = 2×(0) + 3 = 0 + 3 = +3
f(0) = 3
Solve for x = 1, f(1) = 2×(1) + 3 = 2 + 3 = 5
f(1) = 5
Solve for x = 5, f(5) = 2×(5) + 3 = 10 + 3 = 13
f(5) = 13
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Are the given functions y1 and y2 linearly independent or linearly dependent on the interval (0, 1) and why? Choose the correct statement below. Yi = cscº(t) - cot? (t), y2 = 9a. Since yi = - 42 on (0, 1), the functions are linearly independent on (0,1). b. Since yı = 5y2 on (0,1), the functions are linearly independent on (0,1). c. Since yı = 3y2 on (0,1), the functions are linearly dependent on (0,1). d. Since yi = - Žy2 on (0, 1), the functions are linearly dependent on (0,1).
The given functions [tex]y_1[/tex] and [tex]y_2[/tex] are linearly dependent on the interval (0, 1)
(a) Since [tex]y_1 = -42[/tex] on (0,1), the functions are linearly independent on (0,1).
This option does not make sense because [tex]y_1[/tex] is not always -42 on the interval (0,1).
(b) Since [tex]y_1 = 5y_2[/tex] on (0,1), the functions are linearly independent on (0,1).
This option is not correct because if we substitute y1 with 5y2 in the equation [tex]c_1 y_1 + c_2 y_2 = 0[/tex], we get [tex]5c_2 y_2 + c_2 y_2 = 0,[/tex] which means we can write [tex]y_1[/tex]as a constant times [tex]y_2[/tex], so they are dependent.
(c) Since [tex]y_1 = 3y_2[/tex] on (0,1), the functions are linearly dependent on (0,1)."
To see why, we can substitute [tex]y_1[/tex] with [tex]3y_2[/tex] in the equation [tex]c_1 y_1 + c_2 y_2 = 0[/tex]:
[tex]c_1 (3y_2) + c_2 y_2 = 0[/tex]
Simplifying this, we get:
[tex](3c_1 + c_2) y_2 = 0[/tex]
Since [tex]y_2[/tex] is not equal to 0 for any value of t in the interval (0,1), this means that [tex]3c_1 + c_2[/tex] must equal 0. This is a non-trivial solution for [tex]c_1[/tex] and [tex]c_2[/tex], which means that [tex]y_1[/tex] and[tex]y_2[/tex] are linearly dependent on the interval (0,1).
(d) Since[tex]y_1 = -1/2 y_2[/tex]on (0,1), the functions are linearly dependent on (0,1).
This option is also correct because if we substitute [tex]y_1[/tex] with -1/2 y2 in the equation [tex]c_1 y_1 + c_2 y_2 = 0,[/tex] we get [tex]-1/2 c_2 y_2 + c_2 y_2 = 0[/tex], which means we can write[tex]y_1[/tex]as a constant times [tex]y_2[/tex], so they are dependent.
Therefore, the correct answer is either option d, and the functions [tex]y_1[/tex] and [tex]y_2[/tex] are dependent on the interval (0,1).
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Required information M7-20 to 22(Supplement 7A) Calculating Cost of Goods Sold and Ending Inventory under Perpetual FIFO,LIFO,and Weighted Average Cost [LO 7-S1] In its first month of operations,Literacy for the Iliterate opened a new bookstore and bought merchandise in the following order400 units at $5 on January 1,2600 units at $7 on January 8,and 3830 units at $8 on January 29.Assuming 1.030 units are on hand at the end of the month M7-22(Supplement 7A) Calculating Cost of Goods Sold and Ending Inventory under Perpetual Weighted Average Cost [LO 7-S1] Calculate the cost of goods available for sale,cost of goods sold,and ending inventory under the weighted average cost flow assumptions.Assume perpetual inventory system and sold 800 units between January 9 and January 28.(Round your intermediate calculations to 2decimal places.) Weighted Average Cost Goods Avallable for Sale Cost of Goods Sold Ending Inventory Aetivate Window
Calculating Cost of Goods Sold and Ending Inventory under Perpetual FIFO, LIFO, and Weighted Average Cost [LO 7-S1] In its first month of operations, Literacy for the Illiterate opened a new bookstore and bought merchandise in the following order 400 units at $5 on January 1, 2600 units at $7 on January 8, and 3830 units at $8 on January 29.
Assuming 1.030 units are on hand at the end of the month M7-22(Supplement 7A) Calculating Cost of Goods Sold and Ending Inventory under Perpetual Weighted Average Cost [LO 7-S1]
In order to calculate the cost of goods available for sale, cost of goods sold and ending inventory under the weighted average cost flow assumptions, you need to first calculate the weighted average cost of the goods.
Weighted Average Cost = (400 units x $5) + (2600 units x $7) + (3830 units x $8) / (400 + 2600 + 3830)
= $6.62
Cost of Goods Available for Sale = Weighted Average Cost x (400 + 2600 + 3830)
= $6.62 x 7,830
= $51,916.60
Cost of Goods Sold = Weighted Average Cost x 800 units
= $6.62 x 800
= $5,296.00
Ending Inventory = Cost of Goods Available for Sale - Cost of Goods Sold
= $51,916.60 - $5,296.00
= $46,620.60
The cost of Goods Available for Sale is $51,916.60
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Warren is constructing an open box from a piece of paper that is 8 in wide and 11. 5 in long
The sides of the squares be to obtain a box with the largest volume is 114.20 inch³.
Quadratic functions are used in various fields of engineering and science to obtain the values of various parameters. Graphically, they are represented by parabolas. Determines the direction of the curve based on the highest order coefficient. The word "quadratic" is derived from the word "Quad", which means square. In other words, a quadratic function is a "polynomial function of degree 2.
According to the Question:
Let a be the size of the square to be cut at 4 corners.
The cardboard is folded along the dotted line such that following are the box dimensions
Given that
Length, L= 11.5–2a ; Width. B = 8–2a ; Height, H = a
Volume of box, V = (18–2a )(8–2a) a
V = (11.5-2a)(8-2a)a
⇒ V = 92a -3a²-16+ 4a³
⇒ V = 4a³-3a² +92a -16
Now,
The condition for volume to be maximum is that,
dV/da = 0
⇒ [tex]\frac{dv}{da} (4a^3)- \frac{dv}{da} (3a^2) +\frac{dv}{da}(92a) - \frac{dv}{da} (16)[/tex]
⇒ 12a² -6a + 92
Taking common 2 from the above equation, we get:
⇒ 6a² - 3a + 46
Therefore,
The roots of this quadratic equation are,
a = 6.93 and 1.729
Case 1: If a = 6.93,
then B= -5.87 < 0 which is not a valid dimension.
Case 2: If a =1.729 then,
L = 14.54 ; B= 4.54 ; H=1.729
Therefore,
Volume of the box = 14.54 x 4.54 x 1.729 = 114.20
Therefore, the sides of the squares be to obtain a box with the largest volume is 114.20 inch³
Complete Question:
An open box is to be made from 8 inches by 11.5 inches piece of cardboard by cutting squares of equal size from the four corners and bending up the sides. How long should the sides of the squares be to obtain a box with the largest volume?
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Find the distance between the points ( – 9, – 7) and ( – 3,1).
Answer:
10 units
Step-by-step explanation:
calculate the distance d using the distance formula
d = [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }[/tex]
with (x₁, y₁ ) = (- 9, - 7 ) and (x₂, y₂ ) = (- 3, 1 )
d = [tex]\sqrt{(-3-(-9))^2+(1-(-7)^2}[/tex]
= [tex]\sqrt{(-3+9)^2+(1+7)^2}[/tex]
= [tex]\sqrt{6^2+8^2}[/tex]
= [tex]\sqrt{36+64}[/tex]
= [tex]\sqrt{100}[/tex]
= 10 units
[tex] \: [/tex]
[tex] \textrm{( -3 , 1 )}[/tex][tex] \: [/tex]
To find:-[tex] \textrm{Distance between two points = ?}[/tex][tex] \: [/tex]
By using formula:-[tex] \underline{ \star \small \boxed{{ \rm{ \purple{Distance = \sqrt{( x_2 - x_1 )² + ( y_2 - y_1 )²} }}}}}[/tex][tex] \: [/tex]
Solution:-[tex] \rm \: D =\sqrt{ ( x_2 - x_1 )² + ( y_2 - y_1 )²} [/tex][tex] \: [/tex]
where ,
[tex] \: \: \star \rm \red{x_1 = -9 , x_2 = -3 } \\\star \green {\rm y_1 = -7 , y_2 = 1}[/tex]
[tex] \: [/tex]
[tex] \rm \: D = \sqrt{ ( -3 - ( - 9 ))² + ( 1 - (-7 ))²} [/tex][tex] \: [/tex]
[tex] \rm \: D = \sqrt{( -3 + 9)² + ( 1 + 7 )²} [/tex][tex] \: [/tex]
[tex] \rm \: D = \sqrt{( 6 )² + ( 8 )²} [/tex][tex] \: [/tex]
[tex] \rm \: D = \sqrt{36 + 64 } [/tex][tex] \: [/tex]
[tex] \rm \: D = \sqrt{100} [/tex][tex] \: [/tex]
[tex] \boxed{ \textrm{ \color{green}{D = 10 \: }}}[/tex][tex] \: [/tex]
Therefore , the distance between two points is 10 units !
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Shari is drawing triangles that had the same area the base of each triangle varies inversely with the height what are the possible base and height of the second triangle if the first triangles base is 12 and its height is eight find the height of the triangle whose base is 16
If the first triangles base is 12 and its height is eight, the height of the second triangle whose base is 16 is 6.
If the area of all the triangles Shari is drawing is the same, and the base of each triangle varies inversely with the height, then we can use the formula for the area of a triangle:
A = (1/2) * b * h
where A is the area, b is the base, and h is the height.
Since the area is the same for all the triangles, we can set the equation for each triangle equal to a constant, say k:
A = k = (1/2) * b * h
If the base varies inversely with the height, then we can write:
b = k/h
We can substitute this into the equation for the area:
k = (1/2) * (k/h) * h
Simplifying, we get:
k = (1/2) * k
This implies that k = 0, which is not possible since the area of a triangle cannot be zero. Therefore, we conclude that the base of each triangle cannot vary inversely with the height.
For the second part of the question, we are given that the base of the first triangle is 12 and its height is 8. Using the formula for the area of a triangle, we can solve for the area:
A = (1/2) * 12 * 8 = 48
We can then use the same formula to solve for the height of the second triangle whose base is 16:
48 = (1/2) * 16 * h
Simplifying, we get:
h = 6
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Say my population is USA voters. If I interviewed people in front of a major grocery supermarket. Making sure that I selected supermarkets in high median salary areas. What type of sampling best fits my experiment?
The sampling method that best fits your experiment is convenience sampling.
Describe Sampling?Sampling is the process of selecting a subset of individuals or items from a larger population in order to gather information and draw conclusions about the entire population. Sampling is often used in scientific research, market research, and other fields where it is impractical or impossible to measure the entire population.
There are many different types of sampling methods, including random sampling, stratified sampling, cluster sampling, and convenience sampling. Each method has its own advantages and disadvantages, and the choice of sampling method depends on the specific research question and the resources available.
The sampling method that best fits your experiment is convenience sampling. Convenience sampling is a non-probability sampling method that involves selecting individuals who are readily available and accessible to the researcher. In your case, you are selecting individuals who are present in front of a major grocery supermarket, which is a convenient location. Convenience sampling is often used in situations where it is difficult or impractical to obtain a random sample, and is commonly used in market research and pilot studies.
However, it is important to note that convenience sampling has some limitations. Since the sample is not randomly selected, it may not be representative of the population as a whole. In your case, selecting supermarkets in high median salary areas may introduce bias into the sample, as it may not accurately represent the entire population of USA voters. Additionally, the sample size and characteristics may be affected by factors such as time of day, day of the week, and other factors that may affect who is present in front of the supermarket at a given time. Therefore, the results obtained from this sampling method may not be generalizable to the entire population of USA voters.
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five program systems are prepared so that they work independently of each other. each system has a 0.3 chance of detecting an error. find the probability that at least one program system will detect an error. use 4 decimal places.
The probability that at least one program system will detect an error is 0.8319 (approx) or 0.832 (approx).
How to find the probabilityGiven information:
five program systems are prepared so that they work independently of each other. Each system has a 0.3 chance of detecting an error.
Find the probability that at least one program system will detect an error. Use 4 decimal places.The probability of a system detecting an error is 0.3.
The probability of a system not detecting an error is 1 - 0.3 = 0.7.
Probability that none of the five systems detects an error is, P(error not detected in any of the five systems) = P(not detected in 1st) x P(not detected in 2nd) x ... x P(not detected in 5th) = 0.7 x 0.7 x 0.7 x 0.7 x 0.7 = 0.16807.
The probability that at least one system detects an error is, P(at least one system detects an error) = 1 - P(error not detected in any of the five systems) = 1 - 0.16807 = 0.8319 (approx).
Therefore, the probability that at least one program system will detect an error is 0.8319 (approx) or 0.832 (approx).
Hence, the correct option is 0.832.
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If the mean of 8 9 x and 12 is 10 find the range and its coefficient
As per the given mean value, the coefficient of range is 40%.
We know that the mean of these numbers is 10, which means that if we add them all up and divide by 4 (since there are 4 numbers in the set), the result will be 10. Using algebra, we can set up the following equation:
(8 + 9 + x + 12) / 4 = 10
If we simplify this equation, we get:
(29 + x) / 4 = 10
Multiplying both sides by 4, we get:
29 + x = 40
Subtracting 29 from both sides, we get:
x = 11
So we now know that the four numbers are 8, 9, 11, and 12. To find the range, we simply subtract the lowest value (8) from the highest value (12):
12 - 8 = 4
Therefore, the range of the numbers is 4.
To find the coefficient of range, we use the formula:
Coefficient of range = (Range / Mean) x 100%
Plugging in the values we just found, we get:
Coefficient of range = (4 / 10) x 100%
Coefficient of range = 40%
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a relationship that is noted between two or more variables is termed
The relationship that is noted between two or more variables is termed a correlation. Correlation refers to the statistical association or relationship between two or more variables.
It indicates the degree to which the variables are related or how they are related to each other. Correlation can be positive, negative, or zero, depending on the direction and strength of the relationship between the variables.
A positive correlation indicates that as one variable increases, the other variable also increases. A negative correlation indicates that as one variable increases, the other variable decreases. Zero correlation indicates that there is no relationship between the variables.
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an equation of a circle is given by (x+3)^2+(y_9)^2=5^2 apply the distributive property to the square binomials and rearrange the equation so that one side is 0.
The equation of the circle is [tex]x^2 + y^2 + 6x - 18y + 65 = 0[/tex].
Given:
Equation of the circle is [tex](x+3)^2+(y-9)^2=5^2[/tex]
Expand the equation
[tex](x+3)^2 = (x+3)(x+3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9[/tex]
[tex](y-9)^2 = (y-9)(y-9) = y^2 - 9y - 9y + 81 = y^2 - 18y + 81[/tex]
[tex]5^2 = 25[/tex]
Then, substitute the expanded expressions into the equation
[tex](x+3)^2+(y-9)^2=5^2\\(x^2 + 6x + 9) + (y^2 - 18y + 81) = 25\\[/tex]
Simplify and combine like terms
[tex](x^2 + 6x + 9) + (y^2 - 18y + 81) = 25\\x^2 + y^2 + 6x - 18y + 90 = 25[/tex]
Rearrange the equation so that one side is 0
[tex]x^2 + y^2 + 6x - 18y + 90 = 25\\x^2 + y^2 + 6x - 18y + 90 - 25 = 0\\x^2 + y^2 + 6x - 18y + 65 = 0[/tex]
Thus, the equation of a circle [tex](x+3)^2+(y-9)^2=5^2[/tex] can be rearranged using the distributive property to form [tex]x^2 + y^2 + 6x - 18y + 65 = 0[/tex], with one side equaling 0.
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Stanley is making trail mix out of 10 bags of nuts and 20 bags of dried fruits. He want each new portion of trail mix to be identical, containing the same combination of dried fruits with no bags left over. What is the greatest number of portions of trail mix Stanley can make?
To solve this problem, we can use the Greatest Common Factor (GCF) of 10 and 20.
The GCF is the largest number that divides evenly into both 10 and 20. To find the GCF, we can use a factor tree.
We start with 10 and 20 as our starting numbers.
10 = 2 * 5
20 = 2 * 2 * 5
We can see that both 10 and 20 have a factor of 2 and a factor of 5. The Greatest Common Factor between 10 and 20 is 2 * 5, or 10.
Therefore, the greatest number of portions of trail mix Stanley can make is 10.
Which of the options below are inequalities? Choose all of the correct answers. 3h-7-19 h≥14 h=60 3h +12 < 18 20=h+h 8> h
All the equation shown as inequalities:
3h-7-19 ; h≥14 ; h=60 ; 3h +12 < 18 ; 20=h+h ; 8> h
Explain about the inequalities of the numbers?When resolving an inequality, you can do one of the following:
Add the same amount to each side; •Subtract the same amount from each side;Multiply but rather divide both side with a single positive amount. You must flip the inequality sign if you multiply as well as divide each side from a negative number. On a number line, inequalities are represented by making a straight path and designating the end points through an open or closed circle. An empty circle indicates that the value is not included. A closed circle indicates that the value is included.Thus, all the equation shown as inequalities:
3h-7-19 ; h≥14 ; h=60 ; 3h +12 < 18 ; 20=h+h ; 8> h
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The complete question is-
Which of the options below are inequalities? Choose all of the correct answers.
3h-7-19 ; h≥14 ; h=60 ; 3h +12 < 18 ; 20=h+h ; 8> h
Miko bought five 2-litre bottles of water. She poured all the water equally into 15 bowls. What was the volume of water in each bowl?
The volume of water in each bowl is approximately 666.67 ml.
Miko bought five 2-liter bottles of water, which means that she had a total of 10 liters of water. She then poured all the water equally into 15 bowls. To find out the volume of water in each bowl, we need to divide the total volume of water by the number of bowls.
We can start by converting the 10 liters of water into milliliters (ml), which will make it easier to work with. To do this, we multiply 10 by 1000, since there are 1000 ml in one liter. This gives us a total of 10,000 ml of water.
Next, we divide the total volume of water (10,000 ml) by the number of bowls (15) to find the volume of water in each bowl. This can be written as:
Volume of water in each bowl = Total volume of water / Number of bowls
Volume of water in each bowl = 10,000 ml / 15
Volume of water in each bowl = 666.67 ml (rounded to two decimal places)
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solve( 3x^ 2)+2y +4=0
Answer:
Step-by-step explanation:
You can’t solve this equation as none of the numbers have the same coefficient to solve. If you wanted to solve for x and y, you will need two equations as there are two unknown variables in the equation and the only way to solve for x and y is to use simultaneous method which includes two equations.
A man deposited sh. 600,000 to a savings account in a bank that offers an interest rate of 15% p.a. What amount will be on his account after 8months
We can start by calculating the interest earned by the man over 8 months.
First, we need to convert the annual interest rate of 15% to a monthly rate by dividing it by 12:
15% / 12 = 1.25%
Next, we can use the formula for simple interest to calculate the interest earned:
Interest = Principal x Rate x Time
where:
Principal = sh. 600,000 (the amount deposited)
Rate = 1.25% (the monthly interest rate)
Time = 8/12 (8 months is 2/3 of a year)
Plugging in the values, we get:
Interest = 600,000 x 1.25% x 8/12
Interest = 5,000
Therefore, the man will earn sh. 5,000 in interest over 8 months.
To find the total amount in his account after 8 months, we simply add the interest earned to the initial deposit:
Total amount = Principal + Interest
Total amount = 600,000 + 5,000
Total amount = sh. 605,000
Therefore, the amount on the man's account after 8 months will be sh. 605,000.
Find the roots of the polynomial equation.
x^3-x^2+x+39=0
Answer:
-3, 2+3i, and 2-3i.
Step-by-step explanation:
To find the roots of x^3-x^2+x+39=0, we use the Rational Root Theorem and synthetic division to test possible rational roots. We find that -3 is a root, and divide by (x+3) to get the quadratic factor x^2-4x+13=0. Solving this using the quadratic formula gives us the remaining roots of 2+3i and 2-3i. Therefore, the roots of the equation are -3, 2+3i, and 2-3i.
which of the following are presentation methods? (select all that apply)multiple select question. reciprocal training traditional classroom instruction audiovisual training distance learning
There are four presentation methods to choose from. These include reciprocal training, traditional classroom instruction, audiovisual training, and distance learning. The multiple selection question format asks which presentation methods are included.
What is a presentation method?
A presentation method refers to a way to deliver information to a group of individuals. Presentation methods are often used to teach concepts, inform, or persuade the audience. Various presentation methods are available for different types of information, such as traditional classroom instruction, audiovisual training, and distance learning. These methods vary in how information is presented and how participants engage with the material.Presentation methods that are available Reciprocal Training, Reciprocal Training, often known as reciprocal teaching, is a technique that involves individuals exchanging roles as teachers and learners. This method has been shown to be effective in developing students' comprehension of academic material.Traditional Classroom Instruction, Traditional classroom instruction, also known as classroom instruction, is a technique that has been used for centuries. A teacher is in front of a group of students, who are expected to engage with the material and participate in activities.Audiovisual Training Audiovisual training refers to the use of video, audio, and other visual aids in the delivery of instruction. This approach is thought to be more engaging and memorable than other teaching methods.Distance Learning Distance learning, or online learning, is a form of education that is delivered via the internet. This allows individuals to learn from any location, as long as they have access to the internet. This type of education is often more cost-effective than traditional classroom instruction.Therefore, all of the above presentation methods are effective.
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A rabbit is running with a constant speed of 3m/s. Find out the distance covered by rabbit in 10 seconds
The rabbit covers a distance of 30 meters in 10 seconds at a constant speed of 3m/s.
As we know that distance is equal to speed multiplied by time, we can use this formula to calculate the distance covered by the rabbit in 10 seconds. Given that the speed of the rabbit is constant at 3m/s, we can use this value to calculate the distance it covers in 1 second.
So, distance covered by rabbit in 1 second = Speed of rabbit = 3m/s
Now we can use this value to find the distance covered by the rabbit in 10 seconds, which is equal to 10 times the distance covered by the rabbit in 1 second.
Distance covered by rabbit in 10 seconds = 3m/s x 10 seconds
= 30 meters
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In culinary class, you made fudge brownies and peanut butter brownies. Each batch of fudge brownies makes 1 pan. Each batch of peanut butter brownies makes 9 pans. The class made 5 batches and ended up with 29 pans. How many batches of each type of brownie were made?
Answer: 4 batches of fudge brownies and 1 batch of peanut butter brownies were made.
Step-by-step explanation:
Let x be the number of batches of fudge brownies made, and y be the number of batches of peanut butter brownies made.
From the problem, we can write two equations based on the information given:
Each batch of fudge brownies makes 1 pan: x = number of pans of fudge brownies.
Each batch of peanut butter brownies makes 9 pans: 9y = number of pans of peanut butter brownies.
We also know that the class made 5 batches in total, and ended up with 29 pans:
x + 9y = 29 (total number of pans)
We can now solve for x and y by using a system of two equations:
x + 9y = 29 (equation 1)
x + y = 5 (equation 2)
Solving for x in equation 2 and substituting into equation 1, we get:
(5 - y) + 9y = 29
Simplifying and solving for y:
8y = 24
y = 3
Substituting y = 3 into equation 2, we get:
x + 3 = 5
x = 2
Therefore, the class made 2 batches of fudge brownies (2 pans) and 1 batch of peanut butter brownies (9 pans), for a total of 29 pans. Alternatively, we can say that the class made 4 batches of fudge brownies (4 pans) and 1 batch of peanut butter brownies (9 pans) for a total of 29 pans.