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uysha [10]
3 years ago
14

Complete the statement below to explain how this model shows that 1/3÷1/5=5/3

Mathematics
1 answer:
Galina-37 [17]3 years ago
3 0
To divide by a fraction, multiply by the reciprocal of that fraction.

1/3 • 5/1 or 1/3 • 5


Calculate the product straight across.

1/3 • 5/1 = 5/3



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The diagram represents two statements: p and q.
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Answer:

The Region A represents p

The Region B represents p ∧ q

The Region C represents q

Step-by-step explanation:

The figure is as follows :

Given - The diagram represents two statements: p and q.

To find - Which represents regions A, B, and C?

A) p v q

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Solution -

From the figure, we can see that,

The Region A represents p

The Region B represents p ∧ q

The Region C represents q

The truth table is as follows :

p         q        p ∧ q

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5 0
3 years ago
Solve for y. 9x+5y=7 <br>​
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Answer: y=7/5-9x/5

Step-by-step explanation:

im sorry if this dont help

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3 years ago
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Ben says the angle is an acute angle. Jeff says the angle is less than a straight angle. Explain whether their statements are co
Dahasolnce [82]

Answer:

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7 0
2 years ago
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
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