I dont understand this problem

A cone has a volume of 5 cubic inches. What is the volume of a cylinder that the cone fits exactly inside of?

Naya [18.7K]

<span>Volume of Cone: 1/3 Bh where B is the area of the base 1/3 Bh = 5 Bh = 15 B = 15/h The cylinder has the same base area as the cone. Volume of Cylinder: V = B*h B = V/h and B = 15/h The height of the cone and the cylinder are the same. V/h = 15/h Volume of Cylinder = 15.</span>

4 0

2 years ago

A private ship carries two types of cargo: large crates and extra-large crates. Each type of crate weighs a specific amount.

Arisa [49]

Answer:

G. 32

Step-by-step explanation:

32 x 335 = 10720

16,100 - 10720 = 5380/190 = 28.3157894737

275 x 32 = 8800

18,800 - 8800 = 10,000/400 = 25 (This is the closer).

30 x 335 = 10,050

16,100 - 10,050 = 6050/190 = 31.8421052632

30 x 275 = 8250

18,800 - 8250 = 10550/400 = 26.375 (Same reason why it does't work)

25 x 335 = 8375

16,100 - 8375 = 7725/190 = 40.6578947368

25 x 275 = 6875

18,800 - 6875 = 11925/400 = 29.8125

24 x 335 = 8040

16,100 - 8040 = 8060/190 = 42.4210526316

24 x 275 = 6600

18,800 - 6600 = 12200/400 = 30.5

8 0

2 years ago

Mira's kitchen sink holds up to 103.468 liters of water. Round this amount to the nearest liter.

faust18 [17]

103.468

Since 4 rounds down:

103

Hope this helps!

8 0

3 years ago

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The student council is hosting a drawing to raise money for scholarships. They are selling tickets for $9 each and will sell 500

ryzh [129]

Answer:

-$2.32

Step-by-step explanation:

Give that

The selling price per ticket is $9

And the number of tickets sold is 500

So,

There is one grand prize, three $300 second prize, and eleven $40 third prizes

We need to find out the expected value of the profit

$= \frac{1}{500} \times 2000 + \frac{3}{500} \times 300 + \frac{11}{500} \times 40\\\\$

= 4 + 1.8 + 0.88

= 6.68

Now the expected profit is

= $6.68 - $9

= -$2.32

6 0

2 years ago

I own a large truck, and my neighbor owns four small trucks that are all identical. My truck can carry a load of at least $600$

erastova [34]

Answer:

there is no greatest load

Step-by-step explanation:

Let x and y represent the load capacities of my truck and my neighbor's truck, respectively. We are given two relations:

x ≥ y +600 . . . . . my truck can carry at least 600 pounds more

x ≤ (1/3)(4y) . . . . . my truck carries no more than all 4 of hers

Combining these two inequalities, we have ...

4/3y ≥ x ≥ y +600

1/3y ≥ 600 . . . . . . . subtract y

y ≥ 1800 . . . . . . . . multiply by 3

My truck's capacity is greater than 1800 +600 = 2400 pounds. This is a lower limit. The question asks for an <em>upper limit</em>. The given conditions do not place any upper limit on truck capacity.

4 0

2 years ago

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