Find the value of x. Round the length to the nearest tenth.
500 m
A. 777.9 m
40°
Not drawn to scale
B. 595.9 m
C. 321.4 m
D. 652.7 m

Answer:

D={x|x≥2}, R={y|y≥0}.

Step-by-step explanation:

The value of missing side is,

⇒ x = 2.54

We have to given that;

A triangle is shown.

And, To find the missing side of triangle.

Now, By using trigonometry formula, we get;

⇒ tan 23° = Opposite / Base

⇒ tan 23° = x / 6

⇒ 0.424 = x / 6

⇒ x = 0.424 × 6

⇒ x = 2.54

Thus, The value of missing side is,

⇒ x = 2.54

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The side length x=6.63.

By using Pythagoras' theorem on right angle triangle,

"The square of the hypotenuse side is equal to the sum of squares of the other two sides". (Pythagoras' theorem)

(12)^2=(10)^2+(x)^2

(x)^2=144-100

(x)^2=44

The side length is x=6.333.

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A line through points X and Y will be perpendicular bisector of AB.

By examining the figure, it is evident that the compass created arcs as follows:

An arc was drawn with A as the center and a radius greater than half the length of AB.

and, another arc was drawn with B as the center using the same radius.

The points where these arcs intersect are labeled as X and Y.

This construction clearly represents the creation of a perpendicular bisector for the line segment AB.

Therefore, if we draw a line segment passing through points XY, it will serve as the perpendicular bisector of AB.

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The concentration will be 0.5 mg/L at time t =39.86 hours

To determine when the concentration will be 0.5 mg/L, we need to solve the equation:

c(t) = 0.5

20t/(t²+4) = 0.5

Multiplying both sides by (t²+4), we get:

20t = 0.5(t²+4)

Simplifying, we get:

20t = 0.5t² + 2

0.5t² - 20t + 2 = 0

Solving this quadratic equation using the quadratic formula, we get:

[tex]2\left(10+3\sqrt{11}\right)[/tex]

2(10+3(3.31))

2(10+9.93)

39.86

So the concentration will be 0.5 mg/L at t =39.86 hours

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This acquisition of used buses reflects Sheffield Transportation's commitment to providing reliable and cost-effective transportation options for their customers.

In recent years, Sheffield Transportation has acquired three used buses for their transportation fleet. The decision to purchase these buses was likely made to meet the growing demand for transportation services in the area, while also keeping costs down by opting for used vehicles instead of new ones.

It is important to note that when purchasing used buses, Sheffield Transportation would have had to ensure that the vehicles were in good condition and met safety standards before putting them into service. Overall, this acquisition of used buses reflects Sheffield Transportation's commitment to providing reliable and cost-effective transportation options for their customers.

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To solve the inequality 2/3k - 6 < 24, we can start by adding 6 to both sides of the inequality:

2/3k - 6 + 6 < 24 + 6

Simplifying the left side gives:

2/3k < 30

To isolate k, we can multiply both sides by 3/2:

2/3k * 3/2 < 30 * 3/2

Simplifying the left side gives:

k < 45

Therefore, the inequality 2/3k - 6 < 24 expressed in its simplest form is k < 45.

Answer:

Step-by-step explana

To fill in the table, we can use the following formula:

Account value at end of year = (Initial investment + amount borrowed) x (1 + interest rate) x (1 + mutual fund return rate)

Let's first calculate the amount borrowed based on the margin level:

50% margin level: $1,000 x 50% = $500 borrowed

75% margin level: $1,000 x 75% = $750 borrowed

100% margin level: $1,000 x 100% = $1,000 borrowed

Now, let's use the formula to fill in the table:

Margin level Mutual fund return Account value in stellar market Account value in fair market Account value in terrible market

50% 40% $1,500 $1,050 $525

50% 5% $1,100 $1,027.50 $717.45

50% -30% $700 $665 $465

75% 40% $1,750 $1,225 $612.50

75% 5% $1,312.50 $1,221.88 $853.63

75% -30% $875 $831.25 $581.88

100% 40% $2,000 $1,400 $700

100% 5% $1,500 $1,400 $980

100% -30% $1,000 $950 $665

Note that the account value in each market scenario is lower than the initial investment plus the amount borrowed. This is because Ike has to pay back the borrowed amount with interest at the end of the year. The interest rate is 10%, so the account value has to be higher than the initial investment plus the amount borrowed by at least 10% in order to make a profit.tion:

The reduced row echelon form of the matrix is given as follows:

[tex]\left[\begin{array}{cccc}-1&0&0&8\\0&1&0&-2\\0&0&1&-6\end{array}\right][/tex]

The matrix in the context of this problem is defined as follows:

[tex]\left[\begin{array}{cccc}-1&5&-3&6\\0&3&6&-42\\0&-6&-1&18\end{array}\right][/tex]

First we want a value of 1 at line 1, column 1, hence we multiply the first line by -1, that is:

R1 -> -R1.

Hence:

[tex]\left[\begin{array}{cccc}1&-5&3&-6\\0&3&6&-42\\0&-6&-1&18\end{array}\right][/tex]

Then we want a value of 1 at line 2 column 2, thus:

R2 -> R2/3.

Hence:

[tex]\left[\begin{array}{cccc}1&-5&3&-6\\0&1&2&-14\\0&-6&-1&18\end{array}\right][/tex]

Then we want a value of 0 at line 3, column 2, hence:

L3 -> L3 + 6L2

[tex]\left[\begin{array}{cccc}1&-5&3&-6\\0&1&2&-14\\0&0&11&-66\end{array}\right][/tex]

Then the solution to the system of equations is given as follows:

Hence the row echelon form of the matrix is given as follows:

[tex]\left[\begin{array}{cccc}-1&0&0&8\\0&1&0&-2\\0&0&1&-6\end{array}\right][/tex]

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The completed two-column table that we can  use to prove the congruence of ∠BAD and ∠DCB indicates that the missing proof in step 6 of the table is substitution.

Therefore correct option is option B

Congruent angles are described as angles that have the same measurement.

The two-column table is shown below;

 Statements                                                           Reasons

1. ABCD is a parallelogram with diagonal                Given

2. ║ and ║                                                       Definition of a parallelogram

3. ∠2≅∠3, ∠1≅∠4                                              Alt. int. ∠s are ≅

4. m∠2 = m∠3 and m∠1 = m∠4                         Measures of ≅ ∠s are =

5. m∠1 + m∠2 = m∠4 + m∠2                              Addition prop. of equality

6. m∠1 + m∠2 = m∠4 + m∠3                              Substitution

7. m∠1 + m∠2 = m∠BAD                                     Angle addition property

m∠3 + m∠4 = m∠DCB

8. m∠BAD = m∠DCB                                          Substitution

9. ∠BAD ≅ ∠DCB                                              Angles are congruent if   their measures are the same    

Therefore, the missing reason in the partial proof is therefore substitution.

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[tex]\tan(44^o )=\cfrac{\stackrel{opposite}{29}}{\underset{adjacent}{x}} \implies x=\cfrac{29}{\tan(44^o)}\implies x\approx 30.03[/tex]

Make sure your calculator is in Degree mode.

Answer: 98

Step-by-step explanation:

sailboats to dinghies = 5 : 7

if 5 is 70 then what is 7 to ?

well 70/5 is 14 so we had to multiply 5 by 14 to get 70.
That means we also have to multiply 7 by 14 to get 98

The Bakers make 85 glazed donuts.

To solve this problem, we need to determine how many glazed donuts and how many donut holes can be made from 80 pounds of dough. Then, we can compare the quantities of each to determine the difference.

First, we need to convert 80 pounds to ounces, since the amounts of dough for each type of donut are given in ounces. There are 16 ounces in a pound, so 80 pounds is equal to 80 x 16 = 1,280 ounces.

Next, we can set up two equations based on the amounts of dough needed for each type of donut:

Glazed donuts: 10 ounces of dough per donut

Donut holes: 2 ounces of dough per hole

Let's use "g" to represent the number of glazed donuts and "h" to represent the number of donut holes. We know that the total amount of dough used must be 1,280 ounces, so we can write:

10g + 2h = 1,280

We also know that the bakers divide the dough evenly between glazed donuts and donut holes, so the total number of donuts must be:

g + h = ?

We don't know the total number of donuts yet, but we do know that it must be an even number, since the dough is divided evenly between glazed donuts and donut holes.

To solve for g and h, we can use substitution or elimination. Let's use substitution. We can solve the first equation for one of the variables, such as h:

h = (1,280 - 10g) / 2

Then, we can substitute this expression for h in the second equation:

g + (1,280 - 10g) / 2 = ?

Simplifying this equation, we get:

g + 640 - 5g = ?

Combining like terms, we get:

5g + 640 = ?

Subtracting 640 from both sides, we get:

5g = -640

Dividing both sides by 5, we get:

g = -128

This doesn't make sense as a solution, since we can't have negative numbers of donuts. This means there must be an error in our calculations or assumptions.

Let's go back to the second equation:

g + h = ?

We know that the total number of donuts must be an even number, so we can write:g + h = 2n

where n is some positive integer. We can substitute this expression for g + h in the first equation:10g + 2h = 1,280

10g + 2(2n - g) = 1,280

Simplifying this equation, we get 14g - 4n = 1,28

Dividing both sides by 2, we get:7g - 2n = 640

Now we have two equations:7g - 2n = 640

g + h = 2n

We can use substitution or elimination to solve for g and h. Let's use elimination. We can multiply the first equation by 2 and add it to the second equation:14g - 4n = 1,280

2g + 2h = 4n

Multiplying the first equation by 2, we get:28g - 8n = 2,560

Adding this to the second equation, we get:30g = 2,560

Dividing both sides by 30, we get:g = 85.33

This means that the bakers make 85 glazed donuts. To find the number of donut holes, we can substitute this value

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(a) The 99% confidence interval for the population standard deviation at Bank A is 1.33 to 2.16.

(b) The 99% confidence interval for the population standard deviation at Bank B is 1.41 to 2.29.

To find the confidence interval for the population standard deviation, we can use the chi-square distribution. The formula for the chi-square distribution is:

(X²) / (n-1) = s²

where X² is the chi-square value, n is the sample size, and s is the sample standard deviation.

For Bank A:

n = 20

s = 1.252

Using the chi-square distribution table with a degree of freedom of n-1 = 19 and a 99% confidence level, we find the critical values to be 8.907 and 32.852. Substituting these values into the formula above, we get:

(19 x 1.252²) / 32.852 < o² < (19 x 1.252²) / 8.907

1.78 < o² < 4.67

1.33 < o < 2.16

Therefore, the 99% confidence interval for the population standard deviation at Bank A is 1.33 to 2.16.

For Bank B:

n = 20

s = 1.397

Using the chi-square distribution table with a degree of freedom of n-1 = 19 and a 99% confidence level, we find the critical values to be 8.907 and 32.852. Substituting these values into the formula above, we get:

(19 x 1.397²) / 32.852 < o² < (19 x 1.397²) / 8.907

2.00 < o² < 5.24

1.41 < o < 2.29

Therefore, the 99% confidence interval for the population standard deviation at Bank B is 1.41 to 2.29.

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Since the P-value is less than the level of significance (0.0128 < 0.05), we reject the null hypothesis and conclude that there is sufficient evidence to suggest that Gentle Ben has an overall average glucose level higher than 85.

The level of significance is α = 0.05.

The test statistic is:

z = (x - μ) / (σ / √n)

where x is the sample mean,

μ is the population mean,

σ is the population standard deviation, and

n is the sample size.

Plugging in the given values, we get:

z = (93.5 - 85) / (12.5 / √8) ≈ 2.24

Using a standard normal distribution table or calculator, we find that the P-value is approximately 0.0128.

Since the P-value is less than the level of significance (0.0128 < 0.05), we reject the null hypothesis and conclude that there is sufficient evidence to suggest that Gentle Ben has an overall average glucose level higher than 85.

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Answer:

B

Step-by-step explanation:

One half life there will be 50 %

   another half life and there will be 25 %  

        so    TWO half lives * 28.1 yrs/ half life = 56.2 yrs

The value of the side x is 1. 23

To determine the value, we have that the properties of a square are;

From the information given, we have that;

Hypotenuse side = √3

Opposite side = x

Angle = 45 degrees

Using the sine identity, we have that;

sin θ = opposite/hypotenuse

substitute the values, we get;

sin 45 = x/√3

cross multiply the values, we have;

x = 0. 7071× 1. 7321

x = 1. 23

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The volume of the second right triangular prism is 7488.96 cubic meters.

For the second right triangular prism, we have:

Base: A right triangle with base 10 m and height 12 m

Height: 8 m

To find the volume, we can use the same formula as before:

Volume = (1/2) x base x height x altitude

Using the Pythagorean theorem, we can find that the altitude is:

altitude = √(10² + 12²) = √(244) ≈ 15.62

Now we can substitute the given values into the volume formula:

Volume = (1/2) x 10 x 12 x 8 x 15.62

Volume ≈ 7488.96 m^3

Therefore, the volume of the second right triangular prism is 7488.96 cubic meters.

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Answer:

y = 5/11 is y-intercept. slope = -2/11

Step-by-step explanation:

2x + 11y - 5= 0

add 5 to both sides and subtract 2x from both sides

11y = 5 - 2x

to find y-intercept: make x = 0 (because all along the y-axis, x will be 0).

11y = 5 - 2(0)

    = 5

divide by 11

y = 5/11. this is y-intercept.

to find slope (gradient):

11y = 5 - 2x

divide by 11

y = 5/11  - (2/11) x

slope = -2/11

The probability that a male student is taller than 74 inches is 0.3085.

We are given that

σ = 4 inches.

We are also told that 95% of the heights are between 62 inches and 78 inches, which means that 2.5% of the heights are below 62 inches and 2.5% of the heights are above 78 inches.

Using a standard normal distribution table, we can find the z-scores corresponding to these percentages:

The z-score corresponding to the 2.5% below 62 inches is -1.96.

The z-score corresponding to the 2.5% above 78 inches is 1.96.

For the lower limit:

-1.96 = (62 - μ) / 4

For the upper limit:

1.96 = (78 - μ) / 4

Solving for μ in both equations, we get:

μ = 70.08 for the lower limit

μ = 73.92 for the upper limit

Now, the average of these two estimates:

μ ≈ (70.08 + 73.92) / 2 ≈ 72 inches.

To approximate the probability that a male student is taller than 74 inches, we can standardize the height value using the z-score formula:

z = (x - μ) / σ

z = (74 - 72) / 4 = 0.5

As, the probability that a standard normal random variable is greater than 0.5, which is 0.3085.

Therefore, the probability that a male student is taller than 74 inches is 0.3085.

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The  center and variation of the  data set is determined by calculating  the mean and standard deviation.

The time spent volunteering is centered around 7 hours

The values differ from the average by about 4 hours.

A data set is described as a collection of data which corresponds to one or more database tables, where every column of a table represents a particular variable, and each row corresponds to a given record of the data set in question for tabular data sets.

The mean is found as:

Mean = (6 + 5 + 7 + 8 + 9) / 5 = 7 hours

The Range = maximum value - minimum value

Range = 9 - 5 = 4 hours.

The Deviations are  (-1, -2, 0, 1, 2)

Standard deviation = √ [((-1)² + (-2)² + 0² + 1² + 2²) / 5]

Standard deviation = √ [(1 + 4 + 0 + 1 + 4) / 5]

Standard deviation = √2

Standard deviation = 1.4 hours

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The approximate area to the left of z= 2​ is 0.978

From the question, we have the following parameters that can be used in our computation:

The left of z= 2​

The area to the left of z is calculated by calculating the probability that the z-score is less than 2

In other words, this is represented as

Area = (z < 2)

This can then be calculated using a statistical calculator or a table of z-scores,

Using a statistical calculator, we have the area to be

Area =  0.97725

When this value is approximated, we have the approximated area to ve

Area =  0.978

Hence, the area is 0.978

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It is an exercise where the calculation of the circumference of a circle is made from its radius, it is a very common mathematical operation. The circumference is the distance a point on the edge of a circle moves around its center, and the radius is the distance from the center of the circle to its edge.

Circumference = 2πr

Where "r" is the radius of the circle and "π" is a mathematical constant approximately equal to 3.14. Therefore, the circumference of the circle is obtained by multiplying "π" twice by the value of the radius "r".

It is important to remember that the radius and circumference of a circle are related in a directly proportional way: as the radius increases, so does the circumference, and vice versa.

Therefore we apply the formula C =2πr. What we must do is substitute the radius and the value of pi in the formula and solve, then

C =2πr

C = 2 × 3.14 × 19 km

C = 119.32 km

The circumference of the circle can change, varying the value that we give to pi.

Answer:

  150

Step-by-step explanation:

You want the numerical value for log(a, b, c) = (-10, -10, 15) of ...

  [tex]\log\dfrac{\sqrt[3]{c^8}}{a^4b^7}[/tex]

The applicable rule of logarithms are ...

  log(ab) = log(a) +log(b)

  log(a^b) = b·log(a)

  log(a/b) = log(a) -log(b)

The applicable rule for radicals is ...

  [tex]\sqrt[n]{a^m}=a^{\frac{m}{n}}[/tex]

Using the above rules, we can rewrite the logarithm as ...

  [tex]\log\dfrac{\sqrt[3]{c^8}}{a^4b^7}=\dfrac{8}{3}\log(c)-(4\log(a)+7\log(b))\\\\=\dfrac{8}{3}(15)-(4(-10)+7(-10))=40+(4+7)(10)=\boxed{150}[/tex]

The numerical value of the log expression is 150.

<95141404393>

Answer:

no, Irrational

Step-by-step explanation:

because the solution is a non-terminating decimal. 8.246

Answer:

a. The theoretical probability of drawing a spade from a standard deck of cards is 13/52, which reduces to 1/4, or 0.25, or 25%.

b. The theoretical probability of drawing a face card (jack, queen, or king) from a standard deck of cards is 12/52, which reduces to 3/13, or approximately 0.231, or 23.1%.

c. The theoretical probability of drawing a non-face card from a standard deck of cards is 40/52, which reduces to 10/13, or approximately 0.769, or 76.9%.

d. The theoretical probability of drawing a black jack from a standard deck of cards is 2/52, which reduces to 1/26, or approximately 0.038, or 3.8%.

e. The theoretical probability of drawing a red or black card from a standard deck of cards is 52/52, which reduces to 1/1, or 100%.

f. The theoretical probability of drawing a red face card from a standard deck of cards is 6/52, which reduces to 3/26, or approximately 0.115, or 11.5%.