A concert manager counted 600 ticket receipts the day after a concert. The price for a student ticket was $13.50, and the price for an adult ticket was $17.00. The register confirms that $9,587.50 was taken in. How many student tickets and adult tickets were sold?

Answer:

x = 5/2

Step-by-step explanation:

Given that,

[tex] \cfrac{x}{3} = \cfrac{5}{6} [/tex]

Solution:

Multiplying using criss-cross method;

Divide both sides by 6;

Done!

Hence x = 5/2.

Answer:

x = 2.5

Step-by-step explanation:

x = 3*(5/6)

Answer:

5.29

Step-by-step explanation:

8

[tex] \sqrt{8 { = }^{2} { - 6 {?}^{2} }^{2} } [/tex]

Answer: Missing side length (b) = 5.3 units

Step-by-step explanation:

Hii, I'd love to help you out! (:

We are provided with two sides, which are:

Luckily, there's a formula to find the third side!

The Pythagorean Theorem ([tex]\pmb{a^2+b^2=c^2}[/tex]) is the formula that we will use to find the missing side. In this formula "a" and "b" denote the "legs" of the triangle (the two shorter sides) and "c" denotes the hypotenuse (the longest side which is  opposite the right angle (the 90 degree angle)

NB: a and b are interchangeable, because a+b is the exact same thing as b+a; c is always the hypotenuse.

Now it's time for us to stick in the values.

[tex]\bigstar\underbrace{\boxed{\sf a=6;\;c=8}}[/tex]

This is what we obtain upon sticking in the values:

[tex]\tt 6^2+b^2=8^2[/tex]

Now we can solve for b.

[tex]\tt 36+b^2=64}\\b^2=64-36\\b^2=28\\b\approx5.3[/tex]

Thus, the value of "b" is 5.3. (Rounded to the nearest tenth, or one decimal place - one digit after the decimal point)

_____________

Hope I helped! Best wishes.

Reach far. Aim high. Dream big.

_____________

[tex]\underbrace[/tex]

Answer:

x=80.8

Step-by-step explanation:

follow BODMAS

2015-(2015÷5(x-5)÷5)=2014

2015-(2015÷5x-25÷5)=2014

2015-(405/x-5)=2014

2015-(405-5x)=2014

2015-405+5x=2014

1610+5x=2014

5x=2014-1610

(5x=404)1/5

x=80.8

From the slope, it is possible to classify the lines as parallel.

A line can be represented by a linear function. The standard form for the linear equation is: ax+b , for example, y=2x+7. Where:

a= the slope

b=constant term that represents the y-intercept.

The lines are classified as parallel when they have the same slope. You will have perpendicular lines when one line is the negative reciprocal of the second line. If the lines intersect, none of the previous conditions presented are met.

The given equations are:

6x-2y=-2 (1)

y=3x+12 (2)

First, you should write the equation 1 in the standard form.

-2y=-6x-2

-y=-3x-1

y=3x+1

Therefore, the slope for equation 1 is m=3.

After that, you should indicate the slope for equation 2. Thus, m also is 3.

As, both equations present the same slope (m=3). The lines are parallel.

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Answer: The Product(?) of their slopes is 3, so the lines are parallel

Step-by-step explanation:

summary of what the other person said

The sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

For a sector that has a central angle of θ, and a radius of r;

The sector area, and the arc length are:

[tex]A = \frac{\theta}{360} * \pi r^2[/tex] --- sector area

[tex]L = \frac{\theta}{360} * 2\pi r[/tex] ---- arc length

Here, the given parameters are:

Central angle, θ = 160

Radius, r = 5 inches

The sector area is

[tex]A = \frac{\theta}{360} * \pi r^2[/tex]

So, we have:

[tex]A = \frac{160}{360} * \frac{22}{7} * 5^2[/tex]

Evaluate

A = 34.92

The arc length is:

[tex]L = \frac{\theta}{360} * 2\pi r[/tex]

So, we have:

[tex]L = \frac{160}{360} * 2 * \frac{22}{7} * 5[/tex]

L = 13.97

Hence, the sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

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Answer:

3x²

Step-by-step explanation:

0.482 to 1sf is 0.5

61.2x² to one sf is 60x²

√98.01 to 1sf is √100

√100 = 10

0.5 x 60x² = 30x²

30x² / 10 = 3x²

Answer:

in exploratory factor analysis all the measures variables are related to every legend variable but in confirmatory factor analyse CFA research 10 specifically the number of factors required in the data in which measures variable is related with talent variable

Answer:

16 : 19

Step-by-step explanation:

Given weights of the two castings : 32 lbs and 38 lbs respectively.

Now,

Ratio of the weights of the two castings ;

= 32 : 38

= 16 : 19

32:38

16:19

Answer:

this is my answer

The equation is:

(x-5)^2+(y-(-7))^2=36

I hope it helps!

Answer:  [tex]\boldsymbol{(x - 5)^2 + (y + 7)^2 = 36}[/tex]

=========================================

Work Shown:

[tex](h,k) = (5,-7) = \text{center}\\\\r = 6 = \text{radius}\\\\(x - h)^2 + (y - k)^2 = r^2\\\\(x - 5)^2 + (y - (-7))^2 = 6^2\\\\\boldsymbol{(x - 5)^2 + (y + 7)^2 = 36}\\\\[/tex]

Answer:

[tex]9600cm^{2}[/tex]

Step-by-step explanation:

Get the area of the triangle and the rectangle, then subtract the triangle's area for the rectangle's

1) [tex]V=lwh=5(x)(8)=\boxed{40x}[/tex]

2) [tex]V=\pi r^{2}h=(\pi)(3^{2})(8)=\boxed{72}\pi[/tex]

3) [tex]40x=72\pi\\\\x=\frac{72\pi}{40} \approx \boxed{5.7}[/tex]

Let

[tex]I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]tex

Integrate by parts:

[tex]\displaystyle \int u \, dv = uv - \int v \, du[/tex]

with

[tex]u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \cos(2x) \, dx \implies v = \dfrac12 \sin(2x)[/tex]

Then

[tex]\displaystyle I = \frac12 e^{-2x} \sin(2x) + \int e^{-2x} \sin(2x) \, dx + C[/tex]

Integrate by parts again, this time with

[tex]u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \sin(2x) \, dx \implies v = -\dfrac12 \cos(2x)[/tex]

so that

[tex]\displaystyle I = \frac12 e^{-2x} \sin(2x) - \frac12 e^{-2x} \cos(2x) - \int e^{-2x} \cos(2x) \, dx + C\\\\ \implies I = \frac{\sin(2x)-\cos(2x)}{2e^{2x}} - I + C \\\\ \implies 2I = \frac{\sin(2x) - \cos(2x)}{2e^{2x}} + C \\\\ \implies I = \boxed{\frac{\sin(2x) - \cos(2x)}{4e^{2x}} + C}[/tex]

Answer:

See below

Step-by-step explanation:

x2 + 4x + 12 = 0    Complete the square ( take 1/2 of the 'x' coefficient, then square it   and add it to both sides of the equation

x^2 + 4x + 2^2 + 12 = 2^2     Subtract 12 from both sides

x^2 + 4x + 4 = -8

(x+2)^2 = - 8       sqrt both sides

(x+2) =± sqrt (-8)     simplify just a bit

x+2 = ±(2i sqrt2)      subtract 2 from both sides

x =  -2 ± 2i sqrt 2       <===== I do not see that answer listed

Answer: y = -x + 7

Step-by-step explanation:

The slope is [tex]\frac{5-6}{2-1}=-1[/tex], so we know the equation is of the form [tex]y=-x+b[/tex].

Substituting in the coordinates (1,6) to find b,

[tex]6=-1+b\\\\b=7[/tex]

Thus, the equation is y = -x + 7

Answer:

24.012

Step-by-step explanation:

y=kx^2

54=81k

k=54/81

k=0.667

y=kx^2

y=0.667×36

y=24.012

its false ik this because im one of the boyzz

Answer:

Your answer would be A.

Step-by-step explanation:

okay so, with a best fit line, it needs to match up with your points. unless your points go up evenly. its not going to connect to all your points.

Answer:

12.5%

Step-by-step explanation:

To find what percent of 120 15 is, we must find what 15/120 is.

15/120 = 0.125. To convert to a percent, we move the decimal two places to the right. So, the answer is 12.5%.

Brainliest, please :)

Answer:

12.5%

Step-by-step explanation:

Answer:

9

Step-by-step explanation:

The opposite sides of a parallelogram are congruent.

Answer:

9

Step-by-step explanation:

Parallel sides are always equal. :)

The graph of x >2 is attached below

Clearly, the graph of x> 2 could be area beyond 2 on positive X axis

What is graph?

The graph is a demonstration of curves  which gives the relationship between x and y axis.

The question seems to be incomplete, because options are missing.
so the option can be of several graphs.

Thus, The correct Graph of x >2 has been attached below.

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Answer:

1/3 and 1/4

multiplying 1/3 by 1/4 and 1/4 by 1/3 we get

1/3 * 1/4

1 *1 /3*4

1/12

1/4 * 1/3

1*1 / 4*3

1/12

now, both have same denominator.

Answer:

[tex]f(x) = 9x^2+9x - 18/3x + 6\\= 9(x^2 + x - 2)/3(x+2)\\= 3(x^2+x-2)/x+2\\= 3(x+2)(x-1)/x+2\\= 3x - 3[/tex]

so y = 3x - 3

since 3x + 6 is on the bottom,

[tex]3x + 6\neq 0\\3x\neq -6\\x\neq -2[/tex]

so this means it should be the graph of y = 3x - 3, but with a hole at (-2, -9)

The percent increase of women students is 21.13%.

Given that, an area vocational school had an enrollment of 325 men and 123 women.

Percentage is defined as a given part or amount in every hundred. It is a fraction with 100 as the denominator and is represented by the symbol "%".

In 2006, there were 149 women.

Percentage increase = (New value - original value)/Original value × 100

= (149-123)/123 × 100

= 26/123 × 100

= 21.13%

Therefore, the percent increase of women students is 21.13%.

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Answer:

A set is a well-defined collection of distinct mathematical objects.

If the angle 1 and angle 2 are vertical angles, hence they must be equal to each other, therefore m∠1 = m∠2 = 72º

Angles angle are angles that are equal to equal to each other in a line geometry.

Given the following parameters

m∠1 = 72º,

If the angle 1 and angle 2 are vertical angles, hence they must be equal to each other, therefore m∠1 = m∠2 = 72º

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The equation fourth represents the ellipse that has vertices at (1, 0) and (11, 0), and an eccentricity of 3/5 option fourth is correct.

An ellipse is a locus of a point that moves in a plane such that the sum of its distances from the two points called focus adds up to a constant. It is taken from the cone by cutting it at an angle.

We have:

The vertices of the ellipse = (1, 0) and (11, 0)

Eccentricity of ellipse = 3/5

We have equations:

[tex]\rm \dfrac{\left(x-6\right)^{2}}{100}+\dfrac{y^{2}}{64}=1[/tex]

[tex]\rm \dfrac{\left(x+6\right)^{2}}{64}+\dfrac{y^{2}}{100}=1[/tex]
[tex]\rm \dfrac{\left(x+6\right)^{2}}{16}+\dfrac{y^{2}}{25}=1[/tex]

[tex]\rm \dfrac{\left(x-6\right)^{2}}{25}+\dfrac{y^{2}}{16}=1[/tex]

From the fourth equation:

The vertices of the ellipse:

(1, 0) and (11, 0)

The eccentricity:

c²  = a² - b²

c² = 25-16

c = 3

e = c/a = 3/5

Thus, the equation fourth represents the ellipse that has vertices at (1, 0) and (11, 0), and an eccentricity of 3/5 option fourth is correct.

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The mean and the standard deviation are 16.15 and 12.03, respectively

The table of values is given as:

Work hours           Frequency

0 - <10                        45

10 - <20                      33

20 - <30                     20

30 - <40                       7

40 - <50                      8

Start by rewriting the table using the midpoint of each class

x      f

5     45

15    33

25    20

35    7

45    8

The mean is then calculated as:

[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]

This gives

[tex]\bar x = \frac{5 * 45 + 15 * 33 + 25 * 20 35 * 7 + 45 * 8}{45 + 33 + 20 + 7 + 8}[/tex]

Evaluate the sum and the products

[tex]\bar x = \frac{1825}{113}[/tex]

Divide

[tex]\bar x = 16.15[/tex]

The standard deviation is:

[tex]\sigma =\sqrt{\frac{\sum f(x - \bar x)^2}{\sum f}}[/tex]

This gives

[tex]\sigma= \sqrt{\frac{45 * (5 - 16.15)^2 + 33 * (15 - 16.15)^2 + 20 * (25 - 16.15)^2 + 7 * (35 - 16.15)^2 + 8 * (45 - 16.15)^2 }{45 + 33 + 20 + 7 + 8}}[/tex]

Evaluate the sum and the products

[tex]\sigma= \sqrt{\frac{16350.4425}{113}}[/tex]

Divide

[tex]\sigma= \sqrt{144.694181416}[/tex]

Take the square root

[tex]\sigma= 12.03[/tex]

Hence, the mean and the standard deviation are 16.15 and 12.03, respectively

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