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lana66690 [7]
3 years ago
7

If f(x) = 2x + 3 for all values of x, what is the value of f(-3)?​

Mathematics
1 answer:
IrinaK [193]3 years ago
7 0

Answer:

f(-3)= -3

Step-by-step explanation:

We are given the function:

f(x) = 2x+3

and asked to find f(-3). Essentially, we want to find f(x) when x is equal to -3.

Therefore, we can substitute -3 for each x in the function.

f(x)= 2x+3 at x= -3

f(-3)= 2(-3) +3

Solve according to PEMDAS: Parentheses, Exponents, Multiplication, Division, Addition and Subtraction

Multiply 2 and -3.

f(-3) = (2*-3) +3

f(-3)= (-6)+3

Add -6 and 3.

f(-3)= (-6+3)

f(-3)= -3

If f(x)= 2x+3, then<em> f(-3)= -3</em>

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Answer:

The rearrangement can be 45, 987 , 310

Step-by-step explanation:

Here, we want to rearrange the number such that 9 is worth 10 times as what it is worth presently

The value of 9 presently is 90,000

So 10 times as worth will be 10 * 90,000 = 900,000

So we can have the new arrangement as;

45, 987, 310

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Katarina [22]

Answer: it's B

Step-by-step explanation:

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The scale of a model is 0.5in : 6 ft. find the length of the model for an actual length of 204 ft
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Answer:

17 in

Step-by-step explanation:

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Step-by-step explanation:

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Read 2 more answers
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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