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Yuki888 [10]
3 years ago
5

What is the volume of this box?Help me plz ​

Mathematics
2 answers:
Naily [24]3 years ago
7 0

Answer:

5 x 4 x 2 = 40 Cubic units

Step-by-step explanation:

Anestetic [448]3 years ago
5 0

Answer:

40 cubic units

Step-by-step explanation:

5x2x4=40

Hope this helps baiii <3

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You pay 1.5% interest on your credit card bill every month. This month your
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Step-by-step explanation:

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What is 7p-(-5)+(-1)
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Read 2 more answers
The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti
melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that m = 100, \mu = \frac{1}{100} = 0.01

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

P(X > x) = e^{-\mu x}

This is P(X > 190). So

P(X > 190) = e^{-0.01*190} = 0.1496

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

P(X > x) = 0.08

So

e^{-0.01x} = 0.08

\ln{e^{-0.01x}} = \ln{0.08}

-0.01x = \ln{0.08}

x = -\frac{\ln{0.08}}{0.01}

x = 252.6

Capacity of 252.6 cubic feet per second

5 0
3 years ago
Consider the vector b⃗ b→b_vec with length 4.00 mm at an angle 23.5∘∘ north of east. What is the y component bybyb_y of this vec
Stells [14]

Answer:

  • \large\boxed {1.59 mm}

Explanation:

<u>1. Given vector:</u>

  • length: 4.00 mm = magnitude of the vector
  • angle: 23.5º north of east = 23.5º from the x-axys (counterclockwise)

<u>2. y-component</u>

The y-component may be determined using the sine ratio, the angle from the x-axys (counterclockwise direction), and the magnitude of the vector.

  • sine (23.5º) = y-component / magnitude

  • y-component = magnitude × sine (23.5º) = 4.00 mm × sine (23.5º) = 1.59 mm.

  • \large\boxed{y-component = 1.59 mm}

7 0
3 years ago
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