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Ivan
3 years ago
9

It takes 343 inches³ of water to fill an empty cube. If the sides of the cube were tripled, how much water would be needed to fi

ll the new cube?
Mathematics
1 answer:
pochemuha3 years ago
8 0
I believe it would be 9,261 inches of water to fill the new cube.


To find the maximum capacity of a cube you have to multiply s*s*s (each s stands for one of its sides)

The side length happens to be 7. 7 times 7 times 7 is 343. If we triple that length now it’ll be 21. Multiply 21 by 21 by 21 it’ll give you 9,261 inches
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Step-by-step explanation: i guessed plus 19 hours ago

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What is the slope of line PT? PLZ plz HELP
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3 years ago
Please help my grade is a 68 i need a 70 at least
aleksklad [387]

Answer:

10s²-6s+9

Step-by-step explanation:

C = -3 + 5s²

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8 0
2 years ago
An electronics hobbyist has three electronic parts cabinets with two drawers each.
Andrews [41]

Answer:

a) 0.5 = 50% probability that an NPN transistor will be selected.

b) 0.3333 = 33.33% probability that it came from the cabinet that contains both types

c) 66.67% probability that it comes from the cabinet that contains only NPN transistors

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

a) What is the probability that an NPN transistor will be selected?

1/3 probability that the first cabinet is chosen. This cabinet has two transistors, both of which are NPN, so 100% probability of selecting a NPN transistor.

1/3 probability that the second cabinet is chosen. This cabinet has two transistors, both of which are PNP, so 0% probability of selecting a NPN transistor.

1/3 probability that the second cabinet is chosen. This cabinet has two transistors, one of which is NPN, so 50% probability of selecting a NPN transistor.

So

p = \frac{1}{3}*1 + \frac{1}{3}*0 + \frac{1}{3}*0.5 = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = 0.5

0.5 = 50% probability that an NPN transistor will be selected.

b) Given that the hobbyist selects an NPN transistor, what is the probability that it came from the cabinet that contains both types?

Here we use the conditional probability formula.

Event A: NPN transistor

Event B: From the third cabinet.

50% probability that an NPN transistor will be selected, so P(A) = 0.5.

1/6 probability that it is from the third cabinet and NPN, so P(A \cap B) = \frac{1}{6}

The desired probability is:

P(B|A) = \frac{\frac{1}{6}}{0.5} = 0.3333

0.3333 = 33.33% probability that it came from the cabinet that contains both types.

c) Given that an NPN transistor is selected what is the probability that it comes from the cabinet that contains only NPN transistors?

Either it comes from the cabinet with only NPN transistors, or it comes from the cabinet with both types of transistors. The sum of the probabilities of these outcomes is 100%. So

x + 33.33 = 100

x = 66.67

66.67% probability that it comes from the cabinet that contains only NPN transistors

6 0
3 years ago
Tom bought n DVDs for a total cost of 15n - 2 dollars. Which expression represents the cost of each DVD?
Gala2k [10]

A. n ( 15.n - 2 )

eg n = 2

2 ( 15.2 - 2)

2 ( 30 - 2)

2 (18)

36 = 18 + 18

3 0
3 years ago
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