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3 years ago

If f(x)=-x-2 and g(x)=x^2, what is (g°f)(-6)

Mathematics

1 answer:

Leto [7]3 years ago

6 0

$(g \circ f)(x)=(-x-2)^2=x^2+4x+4\\\\
(g\circ f)(-6)=(-6)^2+4\cdot(-6)+4=36-24+4=16$

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Complete the solution of the equation. Find the value of y when x equals -15. -x ‒ 8y = -1

valkas [14]

To complete the solution of - x - 8y = -1, we first need to substitute x for -15.
As so:

-(-15) - 8y = -1
15 - 8y = -1

Next, we can subtract 15 from each side to get:

15 - 15 - 8y = -1 - 15
- 8y = -16

Now, we can divide by -8 on each side to get the y by itself:

-8y / -8 = -16 / -8
y = -16 / -8
y = 2

There we go! <em>y</em> = 2 when x = -15! Hope I could help you out! Have a good one.

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3 years ago

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3 years ago

An automobile manufacturer finds that 1 in every 2500 automobiles produced has a particular manufacturing defect. (a) Use a bin

Advocard [28]

Answer:

a) 0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

b) 0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

Step-by-step explanation:

Binomial 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.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

To use the Poisson approximation for the binomial, we have that:

$\mu = np$

1 in every 2500 automobiles produced has a particular manufacturing defect.

This means that $p = \frac{1}{2500} = 0.0004$

a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 7000 cars.

This is $P(X = 4)$ when $n = 7000$ . So

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

$P(X = 4) = C_{7000,4}.(0.0004)^{4}.(0.9996)^{6996} = 0.1558$

0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

(b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p.

Using the approximation:

$\mu = np = 7000*0.0004 = 2.8$ . So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 4) = \frac{e^{-2.8}*(2.8)^{4}}{(4)!} = 0.1557$

0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

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3 years ago