Ask question

Ask question

All categories

1 year ago

7

An oblique prism has trapezoidal bases and a vertical height of 10 units. An oblique trapezoidal prism is shown. The trapezoid h

as base lengths of x and 2 x, and a height of x. The distance between the 2 trapezoid bases is 20. The vertical height of the prism is 10. Which expression represents the volume of the prism? 10x3 cubic units 15x2 cubic units 20x3 cubic units 30x2 cubic units

1 answer:

denis23 [38]1 year ago

5 0

Answer:

volume of trapezoidal prism = 15x^2 cubic units

Step-by-step explanation:

First, area of the trapezoidal bases.

Parallel sides measure x and 2x, for an average of 1.5x.

Height = x

Area of trapezoidal base = 1.5x*x = 1.5x^2

Volume of prism = area base * height

(length does not matter, height does)

= 1.5x^2 * 10 = 15x^2

You might be interested in

Julio wants to break his school’s scoring record of 864 points during his 24-game basketball season. During the first 8 games of

Scilla [17]

Answer:

x≥38 points

Step-by-step explanation:

Julio wants to break his school’s scoring record of 864 points during his 24-game basketball season. During the first 8 games of the season, he scored a total of 256 points. Which inequality can be used to find x, the number of points Julio must average per game during the rest of the season to break the record?

julio has a 24 game basketball season.

he has played 8, it means there are 16 more games to go

therefore=

he has scored 256 nts, wic means there are still 608 points to go .

864-256=608

x is the points he more score per every game.

16x=608

to break the records ,he must score an extra point 1

so 16x≥608

x≥38

8 0

1 year ago

Read 2 more answers

A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th

QveST [7]

we know that

the volume of a cone is equal to

$V= \frac{1}{3} \pi r^{2}h$

in this problem

the radius is equal to

$r= \frac{112}{2}= 56ft$

1) <u>Find the height of the cone before the storm</u>

Applying the Pythagorean Theorem find the height

$h^{2} = l^{2}-r^{2}$

$l=65 ft$

$h^{2} = 65^{2}-56^{2}$

$h^{2} = 1,089$

$h=33 ft$

2) <u>Find the volume before the storm</u>

$V= \frac{1}{3}*\pi* 56^{2}*33$

$V=34,496\pi\ ft^{3}$

3) <u>Find the volume after the storm</u>

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

$V= \frac{1}{3}*\pi* (56/3)^{2}*11$

$V= 1,277.63\pi\ ft^{3}$

<u>4) Find how this change affect the volume of the pile</u>

Divide the volume after the storm by the volume before the storm

$\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}$

therefore

<u>the answer part a) is</u>

The volume of the pile after the storm is $\frac{1}{27}$ times the original volume

<u>Part b)</u> Estimate the number of lane miles that were covered with salt

5) <u>Find the amount of salt that was used during the storm</u>

$=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}$

6) <u>Find the pounds of road salt used</u>

$104,358.59*80=8,348,687.2\ pounds$

7) <u>Find the number of lane miles that were covered with salt</u>

$8,348,687.2/350=23,853.39 \ lane\ miles$

therefore

<u>the answer part b) is</u>

the number of lane miles that were covered with salt is $23,853.39 \ lane\ miles$

<u>Part c) </u>How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to $1,277.63\pi\ ft^{3}$

$1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}$

8) <u>Find the pounds of road salt </u>

$4,013.79*80=321,103.20\ pounds$

9) <u>Find the number of lane miles </u>

$321,103.20/350=917.44 \ lane\ miles$

therefore

<u>the answer part c) is</u>

the number of lane miles is $917 \ lane\ miles$

7 0

2 years ago

Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th

Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

(a)

The bid should be greater than $10,000 to get accepted by the seller. Let bid x be a continuous random variable that is uniformly distributed between

$10,000 and $15,000

The interval of the accepted bidding is $[ {\rm{\$ 10,000 , \$ 15,000}]$ , where b = $15000 and a = $10000.

The interval of the provided bidding is [$10,000,$12,000]. The probability is calculated as,

$\begin{array}{c}\\P\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{12000}^{15000}\\\end{array}$

$=1- \frac{[15000-12000]}{5000}\\\\=1-0.6\\\\=0.4$

(b) The interval of the accepted bidding is [$10,000,$15,000], where b = $15,000 and a =$10,000. The interval of the given bidding is [$10,000,$14,000].

$\begin{array}{c}\\P\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{14000}^{15000}\\\end{array} P(X14000)$

$=1- \frac{[15000-14000]}{5000}\\\\=1-0.2\\\\=0.8$

(c)

The amount that the customer bid to maximize the probability that the customer is getting the property is calculated as,

The interval of the accepted bidding is [$10,000,$15,000],

where b = $15,000 and a = $10,000. The interval of the given bidding is [$10,000,$15,000].

$\begin{array}{c}\\f\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\\\\{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\\\\{\rm{ = 1}}\\\end{array}$

(d) The amount that the customer bid to maximize the probability that the customer is getting the property is $15,000, set by the seller. Another customer is willing to buy the property at $16,000.The bidding less than $16,000 getting considered as the minimum amount to get the property is $10,000.

The bidding amount less than $16,000 considered by the customers as the minimum amount to get the property is $10,000, and greater than $16,000 will depend on how useful the property is for the customer.

5 0

2 years ago

Which inequality represents the sentence below? Four and nine tenths less than the product of a number and six is no more than e

RUDIKE [14]

Answer:

The answer is C

Step-by-step explanation: Hope this helps!

7 0

1 year ago

Read 2 more answers

Tickets for the baseball games were $2.50 for general admission and 50 cents for kids. If there were six times as many general a

Makovka662 [10]

There was 3000 general admission tickets sold and 500 kid ticket sold.

How did I get this?

First, we need to see what information we have.
$2.50 = General admission tickets = (G)
$0.50 = kids tickets = (K)
There were 6x as many general admission tickets sold as kids. G = 6K

We need two equations:
G = 6K
$2.50G + $.50K = $7750
Since, G = 6K we can substitute that into the 2nd equation.

2.50(6K) + .50K = 7750
Distribute 2.50 into the parenthesis

15K + .50K = 7750
combine like terms

15.50K = 7750
Divide both sides by 15.50, the left side will cancel out.

K = 7750/15.50
K = 500 tickets
So, 500 kid tickets were sold.

Plug K into our first equation (G = 6k)

G = 6*500
G = 3000 tickets

So, 3000 general admission tickets were sold,

Let's check this:

$2.50(3000 tickets) = $7500 (cost of general admission tickets)
$.50(500 tickets) = $250 (cost of general admission tickets)
$7500 + $250 = $7750 (total cost of tickets)

4 0

1 year ago