1 Find the value of x.

i’m like struggling

Answer:

0.005

Step-by-step explanation:

Answer:

if it is perpendicular to eacha other I e 0

11.87 m² metal sheet would be needed to make the base area of tank.

Volume of cylinder, determines how much material it can carry, is determined by the cylinder's volume. A cylinder is a three-dimensional structure having two parallel, identical bases that are congruent.

It is given that capacity of a closed cylindrical vessel of height 2 m is 3080 liters

Let us assume that Radius of cylinder = r

Then Volume of cylinder = π ×r² ×h

= 2π ×r²× m³

1 m³ = 1000 liters

= 2000 π r²  liters

Volume of tank = Capacity

2000 π r² = 3080

=> 2000 × (22/7) × r² = 3080

=> r² = 49/100

=> r = 7/10 m

=> r = 0.7 m

Base Area of tank = TSA = 2πrh + 2πr²

= 2×(22/7)(0.7)×2 + 2×(22/7)×(0.7)²

= 3.0772 +8.792

= 111.87 m²

Hence, 11.87 m² metal sheet would be needed to make it.

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Total seats in plane: 186

108+64+14

5/7 of 14 is: 10

14÷7= 2

2x5= 10

5/16 of 64 is: 20

64÷16=4

5×4= 20

5/9 of 108 is: 60

108÷9= 12

12x5= 60

60+20+10=90

90/186 of seats are being used

simplified: 15/31

No

Answer:

Proofs attached to answer

Step-by-step explanation:

Proofs attached to answer

The probability that x is less than 2 is approximately 0.4866. The probability that x is greater than 3 is approximately 0.3528. The probability that x is less than 5 is approximately 0.6321. The probability that x is between 2 and 5 is approximately 0.1455.

The given probability density function is an exponential distribution with a rate parameter of λ = 1/3. The formula for P(x < x_0) is the cumulative distribution function (CDF) of the exponential distribution, which is given by:

F(x_0) = ∫[0,x_0] f(x) dx = ∫[0,x_0] 1/3 * 4 * e^(-x/3) dx

a. Write the formula for P(x < x_0):

Using integration, we can solve this formula as follows:

F(x_0) = [-4e^(-x/3)] / 3 |[0,x_0]

= [-4e^(-x_0/3) + 4]/3

b. Find P(x < 2):

To find P(x < 2), we simply substitute x_0 = 2 in the above formula:

F(2) = [-4e^(-2/3) + 4]/3

≈ 0.4866

Therefore, the probability that x is less than 2 is approximately 0.4866.

c. Find P(x > 3):

To find P(x > 3), we can use the complement rule and subtract P(x < 3) from 1:

P(x > 3) = 1 - P(x < 3) = 1 - F(3)

= 1 - [-4e^(-1) + 4]/3

≈ 0.3528

Therefore, the probability that x is greater than 3 is approximately 0.3528.

d. Find P(x < 5):

To find P(x < 5), we simply substitute x_0 = 5 in the above formula:

F(5) = [-4e^(-5/3) + 4]/3

≈ 0.6321

Therefore, the probability that x is less than 5 is approximately 0.6321.

e. Find P(2 < x < 5):

To find P(2 < x < 5), we can use the CDF formula to find P(x < 5) and P(x < 2), and then subtract the latter from the former:

P(2 < x < 5) = P(x < 5) - P(x < 2)

= F(5) - F(2)

= [-4e^(-5/3) + 4]/3 - [-4e^(-2/3) + 4]/3

≈ 0.1455

Therefore, the probability that x is between 2 and 5 is approximately 0.1455.

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Answer: x = 6

Step-by-step explanation:

(7x+3)+85+50 = 180

(7x+3)+135 = 180

7x+3 = 180 - 135 = 45

7x = 45-3 = 42

x = 42 / 7 = 6

x = 6

The equation you would use to properly calculate the dollar cost of entering the park and enjoying rides is Total Cost = Entrance Fee + (Number of Rides x Ride Fee).

In this case, Total Cost is the cost of entering the park and enjoying rides, Entrance Fee is the fee for entering the park, Number of Rides is the number of rides you will be taking, and Ride Fee is the fee charged for each ride.

Thus, plugging in the given values, the equation becomes  Total Cost = Entrance Fee + (Number of Rides x Ride Fee).

Therefore, if the Entrance Fee is $ and each ride costs an additional $ , the Total Cost of entering the park and enjoying rides is $ .

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The answer is 250 kilometers. Peter and his mother traveled a total distance of 250 kilometers on their mail delivery.

Distance is a numerical measurement of how far apart two objects or points are. It is a scalar quantity, meaning that it is only expressed as a magnitude, or numerical value, without any direction.

1 centimeter is equal to 0.01 kilometers, so 50 kilometers would be equal to 5,000 centimeters. The total distance between the four cities is 5,000 centimeters, which is equal to 50 kilometers.

Charltc to Thornton is 1,000 centimeters, Thornton to Avon is 1,500 centimeters, Avon to Ashton is 1,000 centimeters, and Ashton to Charltc is 1,500 centimeters. This gives us a total of 5,000 centimeters, or 50 kilometers.

This can be calculated by multiplying the total distance between the four cities (50 kilometers) by the number of times they traveled the route (5 times, since they flew from Charltc to Thornton, then to Avon, then to Ashton, and then back to Charltc). This gives us 250 kilometers, which is the answer to the question.

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Peter and his mother traveled a total distance of 250 kilometers on their mail delivery. The answer is 250 kilometers.

Distance is a numerical measurement of how far apart two objects or points are. It is a scalar quantity, meaning that it is only expressed as a magnitude, or numerical value, without any direction.

1 centimeter =  0.01 kilometers,

so 50 kilometers = 5,000 centimeters.

Charlton to Thornton= 1,000 centimeters,

Thornton to Avon= 1,500 centimeters,

Avon to Ashton= 1,000 centimeters,

and Ashton to Charlton= 1,500 centimeters.

Total distance= 1,000+1,500+1,000+1,500

= 5,000 centimeters, or 50 kilometers.

The total distance between the four cities is 5,000 centimeters, which is equal to 50 kilometers.

This can be calculated by multiplying the total distance between the four cities (50 kilometers) by the number of times they traveled the route (5 times.

=50*5

=250

Since they flew from Charlton to Thornton, then to Avon, then to Ashton, and then back to Charlton.

250 kilometers is the answer to the question.

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Question:

Peter's mother is a pilot. She often makes deliveries near their community. Peter and his mother flew from Charlton to Thornton to make a mail delivery. Then they continued on to Avon and Ashton and returned to Charlton. How many kilometers did Peter and his mother travel in all?

Answer: 4 is the average rate of change

Step-by-step explanation:

The equation to find the average rate of change is:

So f(2)= 11 and f(5)=23 then you plug these numbers in:

[tex]\frac{23-11}{5-2}[/tex] = [tex]\frac{12}{3}[/tex] = 4

Probability of D2 being sunny = 0.78.

On day 1, which is called D1, the probability of sunny is 0.9. It is also given that a sunny day follows a sunny day with 0.8 chance, and a sunny day follows a rainy day with 0.6 chance.

Therefore, we need to find the chances that D2 is sunny.

There are two possibilities for D2: either it can be a sunny day, or it can be a rainy day.

Now, Let us find the probability of D2 being sunny.

We have the following possible cases for D2.

D1 = Sunny; D2 = Sunny

D1 = Sunny; D2 = Rainy

D1 = Rainy; D2 = Sunny

D1 = Rainy; D2 = Rainy

The probability of D1 being sunny is 0.9.

When a sunny day follows a sunny day, the probability is 0.8.

When a sunny day follows a rainy day, the probability is 0.6.

Therefore, the probability of D2 being sunny is given by the formula:

Probability of D2 being sunny = (0.9 × 0.8) + (0.1 × 0.6) = 0.72 + 0.06 = 0.78.

Therefore, the probability that D2 is sunny are 0.78 or 78%.

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The 41st term of the sequence is 121.

In mathematics, a sequence is a list of numbers or objects that follow a certain pattern or rule. A sequence's terms are typically identified by subscripts, like a1, a2, a3,..., an, where n denotes the number of terms in the sequence.

Sequences can be arithmetic, geometric, or neither, depending on terms follow a static difference, constant ratio, or neither of these series, respectively. Algebra uses geometric sequences to represent exponential development or decay whereas arithmetic sequences are frequently employed to model linear connections.

The given sequence is 2 , 5 , 8 , ...

The common difference is:

d = 5 - 2 = 3

The nth term of a sequence is given as:

an = a1 + (n-1)d

Substituting the value we have:

an = 2 + (n-1)3

an = 3n - 1

a41 = 3(41) - 1 = 122 - 1 = 121

Hence, the 41st term of the sequence is 121.

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Answer: There are 10 students who are in both the hockey and football teams.

Step-by-step explanation: We can use the principle of inclusion-exclusion to find the number of students who are in both the hockey and football teams.

The total number of students in both teams is the sum of the number of students in the hockey team and the number of students in the football team, minus the number of students who are in both teams (to avoid double-counting):

Total = Hockey + Football - Both

Substituting the given values, we get:

100 = 40 + 70 - Both

Simplifying, we get:

Both = 40 + 70 - 100

Both = 10

Therefore, there are 10 students who are in both the hockey and football teams.

(a)The distribution of X (the average of the 49 races) follows a normal distribution with mean 146 minutes and standard deviation 11 minutes. (b)The probability of the average of the sample being between 144 and 149 minutes is 0.5854.(c)The 80th percentile for the average of these 49 marathons is 157.2 minutes.(d) The median of the average running times is 146 minutes.

Part(a) The distribution of X (the average of the 49 races) follows a normal distribution with mean 146 minutes and standard deviation 11 minutes. Part (b) The probability that the average of the sample will be between 144 and 149 minutes in these 49 marathons can be calculated using the z-score formula: z = (x - mean)/standard deviation
For x = 144, z = (144 - 146)/11 = -0.18
For x = 149, z = (149 - 146)/11 = 0.27,using the z-score table, the probability of the average of the sample being between 144 and 149 minutes is 0.5854 (0.4026 + 0.1828).

Part (c) The 80th percentile for the average of these 49 marathons can be calculated using the z-score formula: z = (x - mean)/standard deviation, For the 80th percentile, z = 0.84 (from z-score table). Therefore, x = 146 + (0.84 * 11) = 157.2 minutes. Part (d) The median of the average running times is 146 minutes. The median is the midpoint of the data which means half of the data is above the median and half of the data is below the median. Therefore, the median of the average running times is equal to the mean.

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Answer:

Step-by-step explanation:

348.55

The expressions that are equivalent to (x-2)² is x² - 4x + 4. (option B)

Now, let's look at the expression (x-2)². This is a binomial expression that can be simplified by applying the rules of exponents. Specifically, we can expand this expression as follows:

(x-2)² = (x-2) * (x-2)

= x * x - 2 * x - 2 * x + 2 * 2

= x² - 4x + 4

So, the expression (x-2)² is equivalent to x² - 4x + 4.

However, the problem asks us to identify other expressions that are equivalent to (x-2)². To do this, we can use the process of factoring. We know that (x-2)² can be factored as (x-2) * (x-2). Using this factorization, we can rewrite (x-2)² as:

(x-2)² = (x-2) * (x-2)

= (x-2)²

So, (x-2)² is equivalent to itself.

Hence the correct option is (B).

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Complete Question:

Which expressions are equivalent to (x−2)²?

Select the correct choice.

A. (x + 2) (x - 2)

B. x² - 4x + 4

C. x² - 2x + 5

D. x² + x - 2x

In Rosettenville, a suburb of Johannesburg, South Africa, Robert Brozin and Fernando Duarte acquired the Chicken Land restaurant in 1987, launching Nando's.

The eatery was renamed Nando's in honor of Fernando. The restaurant incorporated influences from former Mozambican Portuguese colonists, many of whom had relocated to Johannesburg's southeast after their country gained independence in 1975. Expansion was an essential component of their vision from the beginning. Nando's had already grown from one restaurant in 1987 to four by 1990. It became increasingly difficult to implement a common strategy and decision-making became inefficient as new outlets were maintained as separate businesses.

In 1995, Nando's International Holdings (NIH) was established as a new international holding because managing this growingly complex global structure had become extremely challenging. The South African branch of Nando's Group Holdings (NGH) was successfully listed on the Johannesburg Stock Exchange on April 27, 1997. NGH was 54% owned by NIH, with the remaining 26% available to the general public and former joint venture partners. The main goals of the share offer and listing were to broaden the group's capital base and enable group restructuring.

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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.

Answer:

24 cm (to nearest cm)

Step-by-step explanation:

XLV = extra large tin volume (cm³)

LV = large tin volume (cm³)

XLR = extre large tin radius (cm)

LR = large tin radius (cm) = 20

XLV = 1.75 × LV

Since the tins are geometrically similar cylinders, we can infer that the volumes and radii of the 2 tins are related;

We know the relationship between the volume of the two tins, i.e. the XL tin is 75% greater in volume than the L tin;

This means the volumetric scale factor or multiplier is ×1.75;

Subsequently, we know:

XLV = 1.75 × LV

Similarly, there is a relationship between the radii of the tins;

The relationship is, however, slightly different;

[tex]XLR = (\sqrt[3]{1.75}) \times LR[/tex]

We need to take the cube root of the volumetric scale factor, reason being, the radius is a linear dimension unlike volume;

Easy way to figure this is radius is in cm, volume is in cm³;

So:

XLR = 1.205... × LR

XLR = 1.205... × 20

XLR = 24.101... --> 24 cm (to nearest cm)

Answer: 15

Step-by-step explanation:

13--2 = 13 + 2 = 15

the variance of X is 9. b) Determine E [2X² - 20X +5]:

Using linearity of expectation, we can find E [2X² - 20X +5] as:

E [2X² - 20X +5] = 2E[X²] - 20E[X] + 5

by the question.

The number of ways to select the first 2 white chips is given by:

Number of ways = (6C2) (5C0) (3C0) = 15

The number of ways to select the third chip as blue given that the first 2 chips were white is given by:

Number of ways = (3C1) = 3

Therefore, the probability of selecting the third chip as blue given that the first 2 chips were white is:

P(The third chip is blue the first 2 were white) = Number of ways / Total number of ways = 3 / 350 = 0.0086 (rounded to 4 decimal places)

c) P(The fourth chip is blue the first 2 were white):

The number of ways to select the first 2 white chips is given by:

Number of ways = (6C2) (5C0) (3C0) = 15

The number of ways to select the third chip as non-white given that the first 2 chips were white is given by:

Number of ways = (8C1) = 8

The number of ways to select the fourth chip as blue given that the first 2 chips were white, and the third chip was non-white is given by:

Number of ways = (3C1) = 3

Therefore, the probability of selecting the fourth chip as blue given that the first 2 chips were white is:

P(The fourth chip is blue the first 2 were white) = Number of ways / Total number of ways = 8*3 / 350 = 0.0686 (rounded to 4 decimal places)

Suppose we have a random variable X such that E[X]= 7 and E[X²] =58.

a) Determine the variance of X:

The variance of X is given by:

Var[X] = E[X²] - (E[X]) ²

Substituting the given values, we get:

Var[X] = 58 - (7) ² = 9

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Answer:

Step-by-step explanation:

The analysis of the data to obtain the confidence interval of the difference in the means indicates;

99% confidence interval for [tex]\bar x[/tex]₁ - [tex]\bar x[/tex]₂

The correct options are;

Name of Procedure

Random

10%

Normal/Large Sample

99% CI = (-0.302, 2.301)

Conclude;

We are 99% confident that the interval give in the previous step captures -0.301 to 2.301 = the true difference in mean heart height for all subject like these who listen to love songs versus classical music.

A confidence interval is a range of value that is likely to contain the true value of a population parameter with a certain degree of confidence.

The two-sample t-test can be used to construct the 99% confidence interval as follows;

([tex]\bar x[/tex]₂ - [tex]\bar x[/tex]₁) ± t(α/2, df) × √(s₁²/n₁ + s₂²/n₂)

Where;

[tex]\bar x[/tex]₂ and [tex]\bar x[/tex]₁ = The sample means of the love song and classical music groups

s₁, and s₂ = The sample standard deviations

n₁ and n₂ = The sample sizes

df = The degrees of freedom

t(α/2, df) = The value from the t-distribution table with a significance level of 0.01 and df = n₁ + n₂ - 2

The data indicates;

n₁ = n₂ = 15

[tex]\bar x[/tex]₁ = 5.07, s₁ = 1.63

[tex]\bar x[/tex]₂ = 4.07, s₂ = 1.13

Therefore, we get;

([tex]\bar x[/tex]₂ - [tex]\bar x[/tex]₁) ± t(α/2, df) × √(s₁²/n₁ + s₂²/n₂)

= (5.07 - 4.07) ± t(0.005, 28) × √(1.62²/15 + 1.13²/15)

= 1 ± 2.763 × 0.469

= 1 ± 1.301

The 99% confidence interval for the difference in the true mean heart for subjects who listen to a love song versus classical music is; (-0.301, 2.301).

The correct statement is; 99% confidence interval for [tex]\bar x[/tex]₁ - [tex]\bar x[/tex]₂

The correct statements, placed in the box are;

Name of Procedure;

Two sample interval for [tex]\bar x[/tex]₁ - [tex]\bar x[/tex]₂

Random

The volunteers are randomly selected

The random condition is met

We have a random sample of 15 subjects who listen to a love song

We have a random sample of 15 subjects who listen to classical music

10%

The 10% condition is met

15 < 10% of all subjects like these who listen to love songs

15 < 10% of all subjects like these  who listen to classical music

Normal/Large Sample

The Normal/Large condition is met

The stemplot of the classical music sample data shows no strong skewness or outliers

The stemplot of the love song music sample data shows no strong skewness or outliers

Therefore;

99% CI = (-0.301, 2.301)

Conclude;

We are 99% confident that the interval given in the previous step captures -0.301 to 2.301 = the true difference in mean heart height for all subject like these who listen to love songs versus classical music.

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Answer:

Y intercept is (0,2) Answer D.

Step-by-step explanation:

I included a graph for equation y=2/3 x + 2.

Question: The heights of adult men can be approximated as normal, with a mean of 70 and a standard deviation of 3, the probability that a man is shorter than 65 inches is approximately 0.0475 or 4.75%.

Let X be the height of an adult man, which follows a normal distribution with mean μ = 70 and standard deviation σ = 3. Then, we need to find the probability that a man is shorter than some height, say x₀. We can write this probability as P(X < x₀).To find P(X < x₀), we need to standardize the random variable X by subtracting the mean and dividing by the standard deviation. This yields a new random variable Z with a standard normal distribution. Mathematically, we can write this transformation as:Z = (X - μ) / σwhere Z is the standard normal variable.

Now, we can find P(X < x₀) as:P(X < x₀) = P((X - μ) / σ < (x₀ - μ) / σ) = P(Z < (x₀ - μ) / σ)Here, we use the fact that the probability of a standard normal variable being less than some value z is denoted as P(Z < z), which is available in standard normal tables.

Therefore, to find the probability that a man is shorter than some height x₀, we need to standardize the height x₀ using the mean μ = 70 and the standard deviation σ = 3, and then look up the corresponding probability from the standard normal table.In other words, the probability that a man is shorter than x₀ can be expressed as:P(X < x₀) = P(Z < (x₀ - 70) / 3)We can now use standard normal tables or software to find the probability P(Z < z) for any value z.

For example, if x₀ = 65 (i.e., we want to find the probability that a man is shorter than 65 inches), then we have:z = (65 - 70) / 3 = -1.67Using a standard normal table, we can find that P(Z < -1.67) = 0.0475. Therefore, the probability that a man is shorter than 65 inches is approximately 0.0475 or 4.75%. Thus, P(X < 65) = 0.0475 or 4.75%. Therefore, the probability that a man is shorter than 65 inches is approximately 0.0475 or 4.75%.

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Construct a triangle PQR with sides PQ=8cm, PR=5cm, and QR=6cm, then draw a circle passing through P, Q, and R. This circle is called the circumcircle of triangle PQR.

We draw a line segment PQ = 8 cm long. From point P, we draw a line segment PR = 5 cm long at an angle of 60 degrees to PQ. Then, we draw a line segment QR = 6 cm long joining points Q and R to complete the triangle. Next, we use a compass to draw a circle passing through points P, Q, and R. This circle is called the circumcircle or circumscribed circle of the triangle, which is the unique circle that passes through all three vertices of the triangle. The circumcircle has a special property that its center is equidistant from the three vertices of the triangle.

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The value of the unknown lengths in these similar triangles is FH is 6.67 units and EG is 27 units.

A triangle is a polygon with three sides and three angles. It is a two-dimensional shape that is commonly studied in mathematics, geometry, and other fields. The sum of the angles in a triangle is always 180 degrees, and the lengths of the sides can vary. Triangles can be classified based on the lengths of their sides and the measures of their angles. Common types of triangles include equilateral, isosceles, scalene, acute, right, and obtuse triangles. Triangles have many important properties and are used in various applications, including construction, engineering, and physics.

Here,

1. Let x be the length of FH. We have:

AB/EF = BD/FH

12/8 = 10/x

Cross-multiplying, we get:

12x = 80

x = 80/12

x ≈ 6.67

Therefore, FH ≈ 6.67.

2. Let y be the length of EG. We have:

AC/BD = FH/EG

15/9 = 5/y

Cross-multiplying, we get:

5y = 135

y = 135/5

y ≈ 27

Therefore, EG ≈ 27.

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The system of inequalities to represent the constraints of the situation are x ≤ 260, y ≤ 320, and x + y ≤ 360.

A group of two or more linear inequalities are grouped together and graphed on a coordinate plane to discover the solution that concurrently solves all of the inequalities. Each inequality forms a half-plane on the coordinate plane, and the location where all the half-planes overlap is where the system is solved. Each point inside the feasible area meets all of the system's inequalities. This region is known as the feasible region. To identify the optimum solution given a set of constraints, systems of linear inequalities are frequently utilised in optimization issues.

Let us suppose the number of boards of Mahagony sold = x.

Let us suppose the number of black walnut boards sold = y.

According to the given problem the equation can be set as follows:

x ≤ 260 (the company has 260 boards of mahogany available)

y ≤ 320 (the company has 320 boards of black walnut available)

x + y ≤ 360 (the company expects to sell at most 360 boards of wood)

Hence, the system of inequalities to represent the constraints of the situation are x ≤ 260, y ≤ 320, and x + y ≤ 360.

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