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Alex73 [517]
2 years ago
6

This diagram shows the dimensions of a concrete piece used to build a deck.

Mathematics
1 answer:
Naddika [18.5K]2 years ago
5 0

If you help me with my last question I will help you with this one

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Rational number are ___ natural numbers?<br> Always <br> Sometimes<br> Never
netineya [11]
Rational numbers are Sometimes natural numbers 
hope this helps :D
3 0
2 years ago
Read 2 more answers
What is equivalent to 3x-4y=6
Andreyy89

Answer:

y= (3/4) x - (3/2)

Step-by-step explanation:

3x - 4y = 6 (using algebra add the 4y to the other side & subtract the 6 to the other side >> 3x - 6 = 4y (then divide everything by 4 to get y by itself) >> 3/4x - 6/4 = y (then simplify) >> 3/4x - 3/2 = y

6 0
2 years ago
Thank you to whoever helps me. I'm confused a little bit.
xeze [42]
The answer to this is 392
5 0
3 years ago
Solve y+1=-3/4 (x+2) and y-3= 2/3 (x-4) please show how you solve it
VashaNatasha [74]

Answer:

y = -3/4x-5/2 and y = 2/3x-1/3

Step-by-step explanation:

y+1 = -3/4(x+2)

y+1 = -3/4x-3/2

y = -3/4x-3/2-1

y = -3/4x-5/2

-----

y-3 = 2/3(x-4)

y-3 = 2/3x-8/3

y = 2/3x-8/3+3

y = 2/3x-1/3

7 0
3 years ago
The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine
salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

                                              f(x)=\frac{e^{\frac{-x}{1000} }}{1000}

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

Answer: a. P(x>3000) = 0.5

              b. P(1000<x<2000) = 0.2325

              c. P(x<1000) = 0.6321

              d. 105.4 hours

Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = \int\limits^b_a {P(x)} \, dx

Then, for the electronic component, probability will be:

P(a<x<b) = \int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx

P(a<x<b) = \frac{1000}{1000}.e^{\frac{-x}{1000} }

P(a<x<b) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

a. For a component to last more than 3000 hours:

P(3000<x<∞) = e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = e^{-3}

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}

P(1000<x<2000) = e^{-2}-e^{-1}

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}

P(0<x<1000) = e^{-1}-1

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

0.1 = 1-e^\frac{-x}{1000}

-e^{\frac{-x}{1000} }=-0.9

{\frac{-x}{1000} }=ln0.9

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.

5 0
3 years ago
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