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murzikaleks [220]
2 years ago
14

The surface areas of two similar figures are 25 in2 and 36 in2. If the volume of the smaller figure is 250 in3, what is the volu

me of the larger figure?
a. 360 in3
b. 300 in3
c. 432 in3
d. 145 in3
Mathematics
2 answers:
Anna71 [15]2 years ago
8 0

Answer:360

Step-by-step explanation:

Slav-nsk [51]2 years ago
6 0
Since the figures are similar, we can establish a rule of three as follows.
We know that the area of the smaller figure is 25in^{2}, and its volume is 250in^{3}. We also know that the area of the larger figure is 36in^{2}; since we don't now its volume, lets represent it with X:
\frac{25in^{2}----\ \textgreater \ 250in^{3}}{36in^{2}----\ \textgreater \ Xin^{3}} 
\frac{25}{36} = \frac{250}{X}
X= \frac{(250)(36)}{25}
X=360

We can conclude that the volume of the larger figure is 360in^{3}; therefore, the correct answer is a.
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Denise and Stacey went to a carnival. The admission fee was $6 per person. Each ride at the carnival costs c dollars. The game b
evablogger [386]
The correct answer is D. 14c + 5g + 12.

The entry fee is $6 each. 6 + 6 = 12
Each girl goes on 7 rides which cost c dollars. 7c + 7c = 14c
The number of games both girls play totals 5. And each game costs g dollars. 2g + 3g = 5g.

Add all the costs up and you get 14c + 5g + 12 as their total cost for going to a carnival.
7 0
2 years ago
Betty paints twice as fast as Dan. Working together, Dan and Betty can paint 2, 400 square feet in 4 hours. Another employee, Su
Natalija [7]

Answer:

2670 square feet. Option e.

Step-by-step explanation:

Dan and Betty can paint 2,400 square feet in 4 hours.

They can paint in one hour \frac{2400}{4} = 600 square feet.

Since given that Betty paints twice as fast as Dan. Let us take an equation:

Let Betty = B, Dan = D and Sue = S

B = 2D

4(B+D) = 2400

4B + 4D = 2400

12D = 2400

D = 200 sq. ft.

B = 2D = 400 sq. ft.

Therefore, Dan can paint 200 square feet in 1 hour and Betty paints twice 400 square feet in 1 hour.

Now given three of them can paint 3,600 square feet in 3 hours.

3( B+D+S) = 3600

3B + 3D + 3S = 3600

3(400) + 3(200) + 3(S) = 3600

1200 + 600 + 3S = 3600

S = 600 Sq. ft.

Sue can paint 600 square feet in one hour.

So sue can paint in 4 hours and 27 minutes.

(\frac{4+27}{60}) × 600

= 2670 square feet. Option e.

8 0
2 years ago
Trees in urban areas help keep air fresh by absorbing carbon dioxide. A city has $2100 to spend on planting spruce and maple tre
insens350 [35]
Let x =  the number of spruce trees
Let y =  the number of maple trees

The information given is summarized in the following table.
Number  Cost/tree  Area used   CO₂ absorption
-----------   -------------   ---------------  ----------------------
      x         $30           600 ft²         650 lb/yr
      y         $40           900 ft²         300 lb/yr

The amount available to spend is $2100, therefore
30x + 40y ≤ 2100
or
(3/4)x + y ≤ 52.5                (1)

The land available for planting is 45,000 ft², therefore
600x + 900y ≤ 45000
or
(2/3)x + y ≤ 50                    (2)
     
The amount of CO₂ removed per year is
A = 650x + 300y               (3)

The shaded area in the graph shown below is the solution region.
Optimum values of A occur at the vertices, as shown.
The maximum removal rate occurs at (70, 3.33) at a rate of 46,500 lb/year.
Because we should have an integral number of trees, we should have
70 spruce and 3 maple trees.

Answer: 70  spruce, 3 maple

3 0
2 years ago
Please help! Will give Brainliest :D
gregori [183]

Given that:

The company uses its cargo vans to deliver packages to locations at a 75-mile radius  from the warehouse.

If delivery location is within 8 miles of the delivery boundary, a cargo van will still be used.

To find: The inequality to represent the instances when a vehicle other than a cargo van is used.

Solution:

Let x is the distance of a location from the warehouse.

Cargo van will be used if delivery location is within 8 miles of the delivery boundary.

Minimum distance for delivery = 75-8 = 67 miles.

Maximum distance for delivery = 75+8 = 83 miles.

So, the company  uses cargo vans for any distance in the range 67 miles to 83 miles. So,

67\leq x\leq 83

Therefore, the required inequality is 67\leq x\leq 83.

6 0
2 years ago
Out of six computer chips, two are defective. If two chips are randomly chosen for testing (without replacement), compute the pr
Ratling [72]

Answer:

The probability that of the two chips selected both are defective is 0.1089.

Step-by-step explanation:

Let <em>X</em> = number of defective chips.

It is provided that there are 2 defective chips among 6 chips.

The probability of selecting a defective chip is:

P(X)=p=\frac{2}{6}=0.33

A sample of <em>n</em> = 2 chips are selected.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 2 and <em>p</em> = 0.33.

The probability function of a Binomial distribution is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0, 1, 2, ...

Compute the probability that of the two chips selected both are defective as follows:

P(X=2)={2\choose 2}(0.33)^{2}(1-0.33)^{2-2}=1\times 0.1089\times 1=0.1089

Thus, the probability that of the two chips selected both are defective is 0.1089.

The sample space of selecting two chips is:

S = (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

     (2, 1),  (2, 3), (2, 4), (2, 5), (2, 6)

     (3, 1), (3, 2), (3, 4), (3, 5), (3, 6)

     (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)

     (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)

     (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)

3 0
2 years ago
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