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Luba_88 [7]
2 years ago
8

A candle manufacturer sells cylindrical candles in sets of three. Each candle in the set is a different size. The smallest candl

e has a radius of 0.5 inches and a height of 3 inches. The other two candles are scaled versions of the smallest, with scale factors of 2 and 3. How much wax is needed to create one set of candles?
Mathematics
2 answers:
mafiozo [28]2 years ago
4 0
<span> it is 27 π cubic inches </span>
Margaret [11]2 years ago
3 0

Answer:

The amount of  wax that is needed to create one set of candles is:

                              27π cubic inches.

Step-by-step explanation:

  • The radius(r) of the smallest candle=0.5 inches.

Height(h) of smallest candle= 3 inches.

      Hence, Volume of smallest candle V_1 is:

                 V_1=\pi r^2h

  • Now the medium candle is one which is obtained by taking a scale factor of 2.

  Hence, radius of medium candle=2r

 Height of medium candle=2h

            Hence, Volume of medium candle V_2 is:

       V_2=\pi\times (2r)^2\times (2h)\\\\V_2=8\pi r^2h

  • Similarly the largest candle is one which is obtained by taking a scale factor of 3.

      Hence, radius of largest candle=3r

      Height of largest candle=3h

        Hence, Volume of largest candle V_3 is:

       V_3=\pi\times (3r)^2\times (3h)\\\\V_3=27\pi r^2h

The amount of wax required to create one set of candle is equal to total volume of all the three candles.

\text{Amount\ of\ wax}=V_1+V_2+V_3\\\\\\\text{Amount\ of\ wax}=\pi r^2h+8 \pi r^2h+27\pi r^2 h\\\\\\\text{Amount\ of\ wax}=36\pi r^2h

Now on putting the value of r and h in the expression we get:

\text{Amount\ of\ wax}=36\times \pi\times (0.5)^2\times 3\\\\\\\text{Amount\ of\ wax}=27\pi\ \text{cubic\ inches}

                  Hence, the answer is:

                       27π cubic inches.

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Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 28% below the target pressure.
Alona [7]

Answer:

Step-by-step explanation:

Hello!

The monitoring system warn the driver when the tire pressure of the vehicle is 28% below target pressure.

Be X: target tire pressure of a certain car (pounds per square inch)

a)

X= 28 psi

If the monitoring system will warn the driver when the pressure is 28% below the target pressure: X-0.28X

First step, you have to calculate the 28% of 28psi

28*0.28= 7.84

Second step, is to subtract the calculated 28% to the target pressure:

28 - 7.84= 20.16

The TPMS will trigger a warning at 20.16 psi.

b)

If X~N(μ;σ²)

μ= 28psi (since the average is on target, then the target pressure for the car will be the average value of the distribution)

σ= 3psi

P(X≤20.16)

The standard normal distribution is tabulated. Any value of any random variable X with normal distribution can be "converted" by subtracting the variable from its mean and dividing it by its standard deviation.

So to calculate each of the asked probabilities, you have to first, "transform" the value of the variable to a value of the standard normal distribution Z, then you use the standard normal tables to reach the corresponding probability.

Z= (X-μ)/σ= (20.16-28)/3= -2.61

Now you have to look for the corresponding value of probability using the Z-table. Since the value is negative you have to the use the left entry of the Z-table, in the first column you'll find the integer and first decimal of the value -2.6- and in the first row you'll find the second decimal value -.-1

The value of probability that corresponds to -2.61 is:

P(Z≤-2.61)= 0.005

c)

You have to calculate the probability of inspecting a tire at random and it being inflated within recommended range, symbolically this is:

P(30≤X≤26)= P(X≤30)-P(X≤26)

Calculate both Z values:

Z= (30-28)/3= 0.67

Z= (26-28)/3= -0.67

P(Z≤0.67)-P(Z≤-0.67)= 0.749 - 0.251= 0.498

The probability of the tire being inflated within recommended inflation range is 0.498.

I hope this helps!

5 0
2 years ago
The archway of the main entrance of a university is modeled by the quadratic equation y = -x^2 + 6x. The university is hanging a
Galina-37 [17]
So the problem ask on which of the following choices you had is the points should the banner be attached in the archway and the best answer among the choices will be letter C. (1.5, 4.87) and (3.5, 4.37). I hope this will help you and feel free to ask for more.
8 0
3 years ago
Read 2 more answers
A rectangular prism must have a base with an area of no more than 27 square meters. The width of the base must be 9 meters less
den301095 [7]

Answer:

The maximum height of the prism is 12\ m

Step-by-step explanation:

Let

x------> the height of the prism

we know that

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A\leq 27\ m^{2}

so

L*W\leq 27 -------> inequality A

W=x-9 ------> equation B

L=W+6 -----> equation C

Substitute equation B in equation C

L=(x-9)+6

L=x-3 ------> equation D

Substitute equation B and equation D in the inequality A

(x-3)*(x-9)\leq 27-------> using a graphing tool to solve the inequality

The solution for x is the interval---------->[0,12]

see the attached figure

but remember that

The width of the base must be 9 meters less than the height of the prism

so

the solution for x is the interval ------> (9,12]

The maximum height of the prism is 12\ m

8 1
2 years ago
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A circular platform is to be built in a playground. The center of the structure is required to be equidistant from three support
castortr0y [4]

Answer:

The coordinates for the location of the center of the platform are (0, 1)

Step-by-step explanation:

The equation of the circle of center (h , k) and radius r is:

(x - h)² + (y - k)² = r²

Now,

- The center is equidistant from any point lies on the circumference of the circle

- There are three points equidistant from the center of the circle

- We have three unknowns in the equation of the circle h , k , r

Thus, let's substitute the coordinates of these point in the equation of the circle to find h , k , r.

The equation of the circle is (x - h)² + (y - k)² = r²

∵ Points A(2,−3), B(4,3), and C(−2,5)

- Substitute the values of x and y the coordinates of these points

Point A (2 , -3)

(2 - h)² + (-3 - k)² = r² - - - (1)

Point B (4 , 3)

(4 - h)² + (3 - k)² = r² - - - - (2)

Point C (-2 , 5)

(-2 - h)² + (5 - k)² = r² - - - - (3)

- To find h , k equate equation (1) and (2) and same for equation (2) and (3) because all of them equal r²

Thus;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)² - - - - - (4)

(4 - h)² + (3 - k)² = (-2 - h)² + (5 - k)² - - - - -(5)

- Simplify (5);

h² - 8h + 16 + k² - 6k + 9 = h² + 4h + 4 + k² - 10k + 25

h² and k² will cancel out to give;

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Rearranging, we have;

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(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)²

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h + 3k = 3

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12(3 - 3k) - 4k = -4

36 - 36k - 4k = -4

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5 0
2 years ago
How do i solve this??
mestny [16]

area for the hexagon = 1/2*p*a

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a = apothem = 15.59

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841.86-378 = 463.86 rounded to 464 square units

3 0
2 years ago
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