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Sonja [21]
3 years ago
12

Find the volume of this Composite solid

Mathematics
1 answer:
damaskus [11]3 years ago
5 0

Answer:

D. 569.39 m^3

Step-by-step explanation:

First, we can split up the solid into two shapes: a cone and a cylinder.

Then we can find the volume of the cone

The formula is v= pi r^2 h/3

We know the radius is four and so is the height.

Plug it in and we get 67.02 m^3

Next is the cylinder

Formula is pi r^2 h

We know that the radius is 4 and the height is 10

Plus it in and we get 502.65 m^3

We add 67.02 and 502.65 together to get the total volume.

Answer is D. 569.39 m^3

Mine is a bit higher than D since I used a pi button instead of 3.14

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which equation is the equation of the line in point slope form that has a slope of 12 and passes through the point (-1,14)
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y - 14 = 12( x + 1)

the equation of a line in ' point - slope form ' is

y - b = m(x - a) → (m is slope and (a, b) a point on the line)

here m = 12 and (a , b) = ( - 1, 14)

y - 14 = 12(x - (-1)) ⇒ y - 14 = 12( x + 1)


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3 years ago
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What is the range and domain of the set? :R: {(−11, 16), (17, 23), (2, -13), (7, 4), (-7, −21)}
Ne4ueva [31]

Step-by-step explanation:

the range is the biggest number subtracted by the smallest 23-(-11)= 34

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3 years ago
2. CTfastrak bus waiting times are uniformly distributed from zero to 20 minutes. Find the probability that a randomly selected
Juliette [100K]

Answer:

b. 0.25

c. 0.05

d. 0.05

e. 0.25

Step-by-step explanation:

if the waiting time x follows a uniformly distribution from zero to 20, the probability that a passenger waits exactly x minutes P(x) can be calculated as:

P(x)=\frac{1}{b-a}=\frac{1}{20-0} =0.05

Where a and b are the limits of the distribution and x is a value between a and b. Additionally the probability that a passenger waits x minutes or less P(X<x) is equal to:

P(X

Then, the probability that a randomly selected passenger will wait:

b. Between 5 and 10 minutes.

P(5

c. Exactly 7.5922 minutes

P(7.5922)=0.05

d. Exactly 5 minutes

P(5)=0.05

e. Between 15 and 25 minutes, taking into account that 25 is bigger than 20, the probability that a passenger will wait between 15 and 25 minutes is equal to the probability that a passenger will wait between 15 and 20 minutes. So:

P(15

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how wide can a 5 m load be to fit under a parabolic underpass that is 7 m wide and 7 m at its highest point
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Answer:

\boxed{\text{3.74 m}}

Step-by-step explanation:

This is equivalent to finding the equation of a parabola from three points.

Let's call the ground-level points (-3.5, 0) and (3.5, 0).

The axis of symmetry will be the y-axis, so the vertex is at (0, 7).

Part 1. Find the equation of the parabola

The vertex form of the equation for a parabola is

y = a(x - h)² + k

where (h, k) is the vertex of the parabola.

If the vertex is at (0, 7),  

h = 0 and k = 7

The equation is

y = ax² + 7

Insert the point (0, -3.5)

0 = a(0 + 3.5)² + 7

0 = a(3.5)² + 7

0 = 12.25a + 7

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a = -0.5714

The equation in vertex form is

y = -0.5714x² + 7

Part 2. Calculate the width when y = 5

 5 = -0.5714x² + 7

-2 = -0.5714x²

x² = 2/0.5714

x² = 3.5

x = ±√3.5 = ±1.87  

So, on the graph, the coordinates of the truck at ground level are x = -1.87 and x = +1.87.

The maximum width of the load is 1.87 – (-1.87) =\boxed{\textbf{3.74 m}}

In the diagram below, the red parabola represents the underpass, and the blue box is the truck going through it.

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Sofia has fruit in a basket she has 3 3/8 and 2 7/8 she evenly splits it amount 4 ppl
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the answer for this question 25/16

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