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larisa [96]
3 years ago
13

G(x) = 15 – 4x h(x) = x +8 Write g(h(x)) as an expression in terms of 2.

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
3 0

For this case we have the following functions:

g (x) = 15-4x\\h (x) = x + 8

We must findg (h (x))when x = 2.

So:

g (h (x)) = 15-4 (x + 8) =

We apply distributive property to the terms within parentheses taking into account that:

- * + = -\\15-4x-32 =

We add similar terms taking into account that different signs are subtracted and the sign of the major is placed:

-17-4x

Thus, we have to:

g (h (x)) = - 17-4x

Then, with x = 2:

g (h (2)) = - 17-4 (2) = - 17-8 = -25

Equal signs are added and the same sign is placed.

Answer:

g (h (2)) = - 25

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How many anagrams can be created from the word 'masslessness' if the new words do not need to be meaningful?
Lunna [17]

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4 0
3 years ago
Hope Amelia Solo, the American soccer goalkeeper, World Cup champion and two-time Olympic gold medalist, allows goals at a rate
Nuetrik [128]

Answer:

3.70% probability she allows more than one goal

Step-by-step explanation:

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 time interval.

X~Pois(0.3)

This means that \mu = 0.3

In her next match, what is the probability she allows more than one goal?

Either she allows at most one goal, or she allows more than one goal. The sum of the probabilities of these events is decimal 1. So

P(X \leq 1) + P(X > 1) = 1

We want P(X > 1). So

P(X > 1) = 1 - P(X \leq 1)

In which

P(X \leq 1) = P(X = 0) + P(X = 1)

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

P(X = 0) = \frac{e^{-0.3}*(0.3)^{0}}{(0)!} = 0.7408

P(X = 1) = \frac{e^{-0.3}*(0.3)^{1}}{(1)!} = 0.2222

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.7408 + 0.2222 = 0.9630

P(X > 1) = 1 - P(X \leq 1) = 1 - 0.9630 = 0.0370

3.70% probability she allows more than one goal

3 0
3 years ago
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