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3 years ago

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

1 answer:

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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Rearrange b + 4 = a to b=

mojhsa [17]

$b+4=a\\
b=a-4$

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3 years ago

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The diagram represents two statements: p and q.

suter [353]

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

B) p -> q

C) q ^ p

D) q -> p

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

T T T

T F F

F T F

F F F

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3 years ago

Solve for y. 9x+5y=7 <br>

Yuliya22 [10]

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:

sometimes, if their average angle is over 45 degrees

never are 2 obtuse angles summing to 180 degrees

always. two 90 degree angles always sum to 180 which is a straight line

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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:

$P(X > x) = 0.08$

$e^{-0.01x} = 0.08$

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

$-0.01x = \ln{0.08}$

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

$x = 252.6$

Capacity of 252.6 cubic feet per second

5 0

3 years ago