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melamori03 [73]
3 years ago
5

Which expression is equivalent to -18a^-2b^5/-12a^-4b^-6?

Mathematics
2 answers:
Sonbull [250]3 years ago
8 0

Answer:

\frac{-18a^{-2}b^5}{-12a^{-4}b^{-6}}=\frac{3a^{2}b^{11}}{2}

Step-by-step explanation:

Given : Expression  \frac{-18a^{-2}b^5}{-12a^{-4}b^{-6}}

To find : Which expression is equivalent to given expression?

Solution :

Given expression is

\frac{-18a^{-2}b^5}{-12a^{-4}b^{-6}}

Applying the exponent rule of division, \frac{x^a}{x^b} =x^{a-b}

=\frac{-18a^{-2+4}b^{5+6}}{-12}

=\frac{3a^{2}b^{11}}{2}

Therefore, The required expression is \frac{-18a^{-2}b^5}{-12a^{-4}b^{-6}}=\frac{3a^{2}b^{11}}{2}

Novosadov [1.4K]3 years ago
5 0
First divide the  coefficients:-
-18 / -12  = 3/2

a^-2 / a^-4  = a^2

b^-5 / b^-6  =  b


so the expression simplifies to  3a^2b / 2
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Type the correct answer in the box use/ for the fraction bar
Zarrin [17]

Answer:

see explanation

Step-by-step explanation:

given the 2 equations

y = x² - 2x - 19 → (1)

y + 4x = 5 → (2)

substitute y = x² - 2x - 19 into (2)

x² - 2x - 19 + 4x = 5 ( subtract 5 from both sides )

x² + 2x - 24 = 0 ← in standard form

(x + 6)(x - 4) = 0 ← in factored form

equate each factor to zero and solve for x

x + 6 = 0 ⇒ x = - 6

x - 4 = 0 ⇒ x = 4

substitute each value of x into (1) for corresponding y- coordinate

x = - 6 : y = (- 6)² - 2(- 6) - 19 = 36 + 12 - 19 = 29 ⇒ (- 6, 29)

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the solutions are (- 6, 29), (4, - 11)


6 0
3 years ago
Let x denote the lifetime of a mcchine component with an exponential distribution. The mean time for the component failure is 25
aliina [53]

Answer:

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

Step-by-step explanation:

Exponential distribution:

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

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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 mean time for the component failure is 2500 hours.

This means that m = \frac{2500}, \mu = \frac{1}{2500} = 0.0004

What is the probability that the lifetime exceeds the mean time by more than 1 standard deviations?

The standard deviation of the exponential distribution is the same as the mean, so this is P(X > 5000).

P(X > x) = e^{-0.0004*5000} = 0.1353

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

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zmey [24]
They made $193.80 total.

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SSSSS [86.1K]

Answer: The last line is the answer

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Step-by-step explanation:

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