I am outside the path. I am the diffference of 2 numbers inside the path. I am less than 5. What number am I?

The correct answer is $3000.

It is quite effective to compare two or more values using the division approach. Therefore, it would be accurate to state that a ratio is a comparison or simplification of two quantities of the same type. This relationship illustrates how many times one quantity is equivalent to the other. The ratio may be defined as the number we use to represent one quantity as a percentage of the others.

As given ratio A:B:C = 2:3:5

let, A=2x, B=3x, C=5x

2x=600

x=600/2

x=$300

Total money = 2x+3x+5x

=x(2+3+5)

=x(10)

=300×10

=$3000

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Answer:

A = 80

B = 10

Step-by-step explanation:

a + b = 90

a = 8b

above are  the two equations that we will used to find A and B.

substitute in 8b for a in the first equation.

a + b = 90

8b + b = 90  Combine like terms

9b = 90  Divide both sides by 9

b = 10

If b is 10, than a must be 80.

Using integrals, the probabilities and desired measures are given as follows:

a) P(0 < X) = 0.5.

b) P(0.6 < X) = 0.392.

c) P(-0.5 ≤ X ≤ 0.5) = 0.125.

d) P(X < -2) = 0.

e) P(X < 0 or X > -0.5) = 1.

f) x = 0.965.

The distribution is given by:

f(x) = 1.5x², -1 < x < 1.

The integral of this function will be used to find the probabilities.

In item a, the probability is given by:

[tex]P(0 < X) = \int_{0}^{1} f(x) dx[/tex]

[tex]P(0 < X) = \int_{0}^{1} 1.5x^2 dx[/tex]

[tex]P(0 < X) = 0.5x^3|_{x = 0}^{x = 1}[/tex]

Applying the Fundamental Theorem of Calculus:

P(0 < X) = 0.5(1)³ - 0.5(0)³ = 0.5 - 0 = 0.5.

Item b is similar as item a, just with lower limit 0.6, hence:

P(0.6 < X) = 0.5(1)³ - 0.5(0.6)³ = 0.5 - 0.108 = 0.392.

For item c, the lower limit is -0.5 and the upper limit is of 0.5, hence:

P(-0.5 ≤ X ≤ 0.5) = 0.5(0.5)³ - 0.5(-0.5)³ = 0.125.

For item d, values of -2 and less are not on the range of the distribution, which is between -1 and 1, hence the probability is of 0.

For item e, the intersection of the intervals is the entire range of the function, hence the probability is of 1.

For item f, we have that:

0.5(1)³ - 0.5(x)³ = 0.05

0.5x³ = 0.45

x³ = 0.9.

x = (0.9)^(1/3) -> cubic root

x = 0.965.

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The common side length of the paintings is 8 feet

The area of a shape is the amount of space on the shape

The given parameters are

Area of painting 1 = 24 square feet

Area of painting 2 = 32 square feet

Express the areas as the product of their factors

So, we have

24 = 2 * 2 * 2 * 3

32 = 2 * 2 * 2 * 4

Multiply the common factors

Common factors =2 * 2 * 2

Evaluate the product

Common factors = 8

This represents the common side length

Hence, the common side length of the paintings is 8 feet

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Answer:

well wheren is the question

Answer:

B

Step-by-step explanation:

3/20 15%,20%, 25%, 1/3 33%

All my work and answer is provided in the attached screenshot to my answer! :)

Have a great day!

[tex]9x = 81+4[/tex]

[tex]9x = 85[/tex]

[tex]x = 85/9[/tex](divide using calculator)

The volume of Daves storage box is 19200 and the surface area is 4400

The given parameters are

Length = 32 cm

Width = 20 cm

Height = 30 cm

The volume is

Volume = Length * width * height

So, we have

Volume = 32 * 20 * 30

Evaluate

Volume = 19200

The surface area is

A =2 *(LW + LH + WHW)

So, we have

A = 2 * (32 * 20 + 20 * 30 + 32 * 30)

Evaluate

A = 4400

Hence, the volume of Daves storage box is 19200 and the surface area is 4400

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Answer:

Step-by-step explanation:

34 with a remainder of 5. so basically you can say 34 1/2

Answer:

350

Step-by-step explanation:

Rounding 345=350

345 to the nearest thousands is 0

345 to the nearest hundreds is 300

345 to the nearest tens is 350

345 to the nearest whole number is 345

345 to the nearest tenths is 345.0

The value of the line segment TU is 12 units.

A line segment is a fixed-length section of the a line to two ends.

According to the question;

TU = x + 8 and ST = 3x are the lengths.

SU is the total length of the line segment.

Now, it is clear that SU is divided into parts such that;

SU = ST + TU

As, T is the mid point of the line SU;

The,

ST = TU

Equating the values;

3x = x + 8

Solving the equation;

3x -x = 8

x = 4

Put the value of x in TU;

TU = x + 8

TU = 4 + 8

TU = 12

Therefore, the value of the line segment TU is found as 12 units.

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A light bulb will consume 3,600 watts - hours in 1 day if it consumes 12,600 watts in 3 days 12 hours.

We are given that:

Consumption in 3 days 12 hours = 12,600 watts

12 hours = 12 / 24 days = 0.5 days

Days = 3 + 0.5 days = 3.5 days

Now, by using the unitary method, we get that the consumption by a light bulb in 1 day will be equal to:

3.5 days = 12,600 watts

1 day = 12,600 / 3.5 watts

1 day = 3,600 watts

Therefore, a light bulb will consume 3,600 watt - hours in 1 day if it consumes 12,600 watts in 3 days 12 hours.

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The square's diagonal length is (E) d = 5√2.

So, we must determine the length of the square's diagonal.

Therefore, the square's diagonal length is (E) d = 5√2.

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The correct question is given below:
What is the length of the diagonal of the square shown below? 5 45° 5 5 90° 5 O A. 5

B. 10

C. 5

D. 5/5

E. 5√2

F. 25​

Answer:

x = 6

Step-by-step explanation:

As stated, the scale factor from two figures is 1/4

To find the value of x we can simply use the following equation

[tex] \frac{(1.5)}{x} = \frac{1}{4} [/tex]

cross multiply expressions

x = 6

Answer:

C) Positively skewed, and the mean is to the right of the median.

Step-by-step explanation:

A histogram is a graphical representation of the distribution of numerical data.

A histogram is:

Since the long tail of the given histogram is on the positive side of the peak, the histogram is positively skewed (right-skewed).

The mode is the value that occurs most often in a set of data.

In a histogram, the mode is the highest point.

The median and mean fall to the right of the mode in a right-skewed (positively skewed) histogram, and the mean is always to the right of the median: mode < median < mean.

The median and mean fall to the left of the mode in a left-skewed (negatively-skewed) histogram, and the mean is always to the left of the median: mean < median < mode.

So the correct definition of the given histogram is:

Answer:

15 + (n + 12)

Step-by-step explanation:

The equation that shows the relation between the number of hours worked and tickets sold is t = 23h (fourth option)

Examining the table, it can be seen that the number of tickets sold increases by 23 for every hour that Gwen works.

Tickets sold when Gwen works for 2 hours = 23 x 2 = 46

Tickets sold when Gwen works for 2 hours = 23 x 3 = 69

Because the tickets sold increase by a constant number, the equation would be modelled as a linear function.

Linear equations have the form : a + bx

Ticket sold = (number of hours x ticket sold in the first hour)

t = 23h

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Answer:O 18+3

O-18+3

O-18+(-3)

Step-by-step explanation:O 18+3

O-18+3

O-18+(-3)

Answer:

Step-by-step explanation:

[tex]A=64\pi ft^2[/tex]

[tex]r=\sqrt{\frac{A}{\pi } }[/tex]

[tex]r=\sqrt{\frac{64\pi }{\pi } }[/tex]

[tex]r=\sqrt{64}[/tex]

[tex]r=8[/tex]

So the radius is 8 ft.

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Answer:

4,000

Step-by-step explanation:

Answer:

5

Step-by-step explanation:

By using the definition of perimeter and addition and subtraction of sides, the perimeter of the composite figure is 22 centimeters.

According to the image attached aside, we find a composite figure created by adding three quadrilaterals. The perimeter is the sum of the lengths of all sides of the composite figure, that is, we need to add the lengths of the eight sides of the figure.

p = AB + BC + CD + DE + EF + FG + GH + AH

GH = AB - EF - CD

GH = 7 cm - 3 cm - 1 cm

GH = 3 cm

AH = BC - DE + FG

AH = 3 cm - 2 cm + 1 cm

AH = 2 cm

p = 7 cm + 3 cm + 1 cm + 2 cm + 3 cm + 1 cm + 3 cm + 2 cm

p = 22 cm

The perimeter of the composite figure is 22 centimeters.

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Based on the fact that Lines a and m are parallel, the degree measure of angle 3 is 127

Based on the fact that a and m are parallel lines, then this means that the angle 127 degrees is corresponding to angle 1.

This means that the measure of angle 1 is 127 degrees as well because corresponding angles are congruent.

Angle 1 and Angle 3 are vertical angles. Vertical angles are also congruent. This means that if Angle 1 is 127 degrees then angle 3 will be 127 degrees as well.

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The length of the side x of the triangle ADE is 16.5 units.

If the sides are in the same ratio or proportion and the angles are equal (corresponding angles), two triangles will be similar (corresponding sides). When compared individually, similar triangles may have different side lengths, but they must all have the same ratio of their side lengths and equal angles.

From figure, it is given that:

AB = 3 units

BC = 2 units

AD = x units

DE = 11 units

We know that the triangle ABC ~ ADE.

Therefore, the ratio of corresponding sides of the triangles will be same.

Now,

AB/ AD = BC/ DE

3/x = 2/11

Cross multiply to get:

2x = 3(11)

2x = 33

x = 33/2

x = 16.5 units

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Answer:

1hour 15

Step-by-step explanation:

The glider begins its flight a mile above the ground.

Distance above the ground after 45 minutes =

Change in height of the glider

Next, we determine how long the entire descent last.

Expressing the distance moved as a ratio of time taken

Therefore: Total Time taken =45+30=75 Minutes

=1 hour 15 Minutes

Using multiplication, Correct option is B. The amount of solution that can be made from 15 L of concentrate is 16.5L.

Multiplying in math is the same as adding equal groups. The number of items in the group grows as we multiply. Parts of a multiplication issue include the product, the two factors, and the product. The components in the multiplication problem 6 x 9 = 54 are the numbers 6 and 9, and the result is the number 54.

Given,

Total ant killers added in the water is 5 × 20mL = 100mL = 0.1L (using 1L = 1000mL)

Total quantity of solution is 0.1L + 1L = 1.1L

It takes 0.1 mL of ant killer to make 1.1L of solution.

⇒ Total solution that can be made from 15L of concentrate is -

⇒ 15L × 0.1L = 16.5L

∴ The solution that can be made from 15L of concentrate is  16.5L

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the domain of the function y = 56.35 x is 0.27 ≤ x ≤ 0.44.

We are given a function:

y = 56.35 x

We are also given that at least 15 but no more than 25 students go to the water park.

This means that the range of y is:

Range = 15 ≤ y ≤ 25.

Now, we need to find the domain of the function.

Put y = 15 in the equation:

15 = 56.35 x

x = 15 / 56.35

x = 0.27

Put y = 25 in the equation:

25 = 56.35 x

x = 25 / 56.35

x = 0.44

So, the domain becomes:

Domain = 0.27 ≤ x ≤ 0.44.

Therefore, we get that, the domain of the function y = 56.35 x is 0.27 ≤ x ≤ 0.44.

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The answer to the given question is length is 17m and width is 6m

Given that a rectangle's length is 7 meters less than its width.

And also given the perimeter of the rectangle is 46 meters.

We know that the rectangle's perimeter is twice the sum of its length and width.

l - length of the rectangle

w - width of the rectangle

P - perimeter of the rectangle

Then we have P = 2 × (l + w)

Given that a rectangle's length is 7 meters less than its width.

⇒ l = 4 × w - 7

Given the perimeter of the rectangle is 46 centimeters.

⇒ 2 × ( l + w) = 46

⇒ 2 × (4w - 7 + w) = 46 (using the value of length given in terms of width)

⇒ 4w - 7 + w = 23 (dividing by 2 on both the sides)

⇒ 5w - 7 = 23

⇒ 5w = 23 + 7

⇒ 5w = 30

⇒ w = 6 (dividing both sides by 5)

Using the width value in the length equation, we get

l = 4 × 6 - 7

⇒ l = 24 - 7

⇒ l = 17

Therefore, The rectangle is 17 meters in length and 6 meters in width.

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Answer: the answer is 42.5

Observe that

[tex]\dfrac{(n!)^2}{(2n+1)!} = \dfrac{n!(2n-n)!}{(2n+1)(2n)!} = \dfrac1{(2n+1)\binom{2n}n}[/tex]

Starting with a well-known series

[tex]\displaystyle 2\arcsin^2(x) = \sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2 \binom{2n}n}[/tex]

we take some (anti)derivatives to find a sum that more closely resembles ours.

Let [tex]f(x)=2\arcsin^2(x)[/tex]. Then

[tex]\displaystyle f'(x) = 2 \sum_{n=1}^\infty \frac{2^{2n} x^{2n-1}}{n \binom{2n}n}[/tex]

[tex]\displaystyle x f'(x) = 2 \sum_{n=1}^\infty \frac{2^{2n} x^{2n}}{n \binom{2n}n}[/tex]

[tex]\displaystyle x f''(x) + f'(x) = 4 \sum_{n=1}^\infty \frac{2^{2n} x^{2n-1}}{\binom{2n}n}[/tex]

[tex]\displaystyle x^2 f''(x) + x f'(x) = 4 \sum_{n=1}^\infty \frac{2^{2n} x^{2n}}{\binom{2n}n}[/tex]

Noting that both sides go to zero as [tex]x\to0[/tex], by the fundamental theorem of calculus we have

[tex]\displaystyle \sum_{n=1}^\infty \frac{2^{2n} x^{2n+1}}{(2n+1)\binom{2n}n} = \frac14 \int_0^x (t^2 f''(t) + t{}f'(t)) \, dt[/tex]

so that when [tex]x=\frac12[/tex], and rearranging some factors and introducing a constant, we recover a useful sum.

[tex]\displaystyle \sum_{n=0}^\infty \frac1{(2n+1)\binom{2n}n} = 1 + \frac12 \int_0^{1/2} (x^2 f''(x) + x f'(x)) \, dt[/tex]

Integrate by parts.

[tex]\displaystyle \int_0^{1/2} x^2 f''(x) \, dx = \frac14 f'\left(\frac12\right) - 2 \int_0^{1/2} x f'(x) \, dx[/tex]

[tex]\displaystyle \int_0^{1/2} x f'(x) \, dx = \frac12 f\left(\frac12\right) - \int_0^{1/2} f(x) \, dx[/tex]

Then our sum is equivalent to

[tex]\displaystyle \sum_{n=0}^\infty \frac1{(2n+1)\binom{2n}n} = 1 + \frac18 f'\left(\frac12\right) - \frac14 f\left(\frac12\right) + \int_0^{1/2} \arcsin^2(x) \, dx[/tex]

The remaining integral is fairly simple. Substitute and integrate by parts.

[tex]\displaystyle \int_0^{1/2} \arcsin^2(x) \, dx = \int_0^{\pi/6} u^2 \cos(u) \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{\pi^2}{72} - 2 \int_0^{\pi/6} u \sin(u) \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{\pi^2}{72} + \frac\pi{2\sqrt3} - 2 \int_0^{\pi/6} \cos(u) \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{\pi^2}{72} + \frac\pi{2\sqrt3} - 1[/tex]

Together with

[tex]f\left(\dfrac12\right) = 2 \arcsin^2\left(\dfrac12\right) = \dfrac{\pi^2}{18}[/tex]

[tex]f'\left(\dfrac12\right) = \dfrac{4\arcsin\left(\frac12\right)}{\sqrt{1-\frac1{2^2}}} = \dfrac{4\pi}{3\sqrt3}[/tex]

we conclude that

[tex]\displaystyle \sum_{n=0}^\infty \frac{(n!)^2}{(2n+1)!} = \sum_{n=0}^\infty \frac1{(2n+1)\binom{2n}n} \\\\ ~~~~~~~~~~~~~~~~~~ = 1 + \left(\frac18\cdot\frac{4\pi}{3\sqrt3}\right) - \left(\frac14\cdot\frac{\pi^2}{18}\right) + \left(\frac{\pi^2}{72} + \frac\pi{2\sqrt3} - 1\right) \\\\ ~~~~~~~~~~~~~~~~~~ = \boxed{\frac{2\pi}{3\sqrt3}}[/tex]