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2 years ago

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Consider the function represented by the equation 6q = 3s - 9. Write the equation in function notation, where q is the independe

nt variable.
f(q) = 2s + 3

f(q) = 2q + 3

1 answer:

Tom [10]2 years ago

7 0

Answer: The correct option is (B) f(q) = 2q + 3.

Step-by-step explanation: The given equation representing a function is

$6q=3s-9~~~~~~~~~~~~(i)$

We are to write the equation in function notation with 'q' as independent variable.

From equation (i), we have

$6q=3s-9\\\\\Rightarrow 3s=6q+9\\\\\Rightarrow s=2q+3.$

So, 's' can be represented as a function of 'q'. Hence, we can write

$s=f(q).$

Therefore, we have

$f(q)=2q+3.$

Thus, (B) is the correct option.

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Sladkaya [172]

Answer:

a) 0.1423

b) 0.2977

c) 0.56

Step-by-step explanation:

For each driver stopped for speeding, there are only two possible outcomes. Either they have invalid licenses, or they do not. The probability of a driver having an invalid license is independent from other drivers. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

13 percent of the drivers stopped for speeding have invalid licenses.

This means that $p = 0.13$

14 drivers are stopped

This means that $n = 14$

(a) None will have an invalid license.

This is $P(X = 0)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.13)^{0}.(0.87)^{14} = 0.1423$

(b) Exactly one will have an invalid license.

This is $P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{14,1}.(0.13)^{1}.(0.87)^{13} = 0.2977$

(c) At least 2 will have invalid licenses.

Either less than 2 have invalid licenses, or at least 2 does. The sum of the probabilities of these events is decimal 1. Mathematically, this is

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

So

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.1423 + 0.2977 = 0.44$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.44 = 0.56$

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2 years ago

i’m order for a graph to be proportional it must cross through the origin. A.true B.false

Gelneren [198K]

Answer:

True

Step-by-step explanation:

Key idea: the graph of a proportional relationship is a straight line through the origin.

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Jerome hiked three fourths of a mile in two thirds of an hour. How many miles can Jerome hike in one hour?

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3 years ago

Please help I will give Brainliest please!

WITCHER [35]

Part (a)

The domain is the set of allowed x inputs of a function.

The graph shows that x = 0 is not allowed because of the vertical asymptote located here. It seems like any other x value is fine though.

<h3>Domain: set of all real numbers but $x \ne 0$ </h3>

To write this in interval notation, we can say $(-\infty, 0) \cup (0, \infty)$ which is the result of poking a hole at 0 on the real number line.

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The range deals with the y values. The graph makes it seem like it stretches on forever in both up and down directions. If this is the case, then the range is the set of all real numbers.

<h3>Range: Set of all real numbers</h3>

In interval notation, we would say $(-\infty, \infty)$ which is almost identical to the interval notation of the domain, except this time of course we aren't poking at hole at 0.

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Part (b)

<h3>The x intercepts are x = -4 and x = 4</h3>

We can compact that to the notation $x = \pm 4$

These are the locations where the blue hyperbolic curve crosses the x axis.

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Part (c)

<h3>Answer: There aren't any horizontal asymptotes in this graph.</h3>

Reason: The presence of an oblique asymptote cancels out any potential for a horizontal asymptote.

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Part (d)

The vertical asymptote is located at x = 0, so the equation of the vertical asymptote is naturally x = 0. Every point on the vertical dashed line has an x coordinate of zero. The y coordinate can be anything you want.

<h3>Answer: x = 0 is the vertical asymptote</h3>

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Part (e)

The oblique or slant asymptote is the diagonal dashed line.

It goes through (0,0) and (2,6)

The equation of the line through those points is y = 3x

If you were to zoom out on the graph (if possible), then you should notice the branches of the hyperbola stretch forever upward but they slowly should approach the "fencing" that is y = 3x. The same goes for the vertical asymptote as well of course.

<h3>Answer: Oblique asymptote is y = 3x</h3>

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2 years ago