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Anna35 [415]
3 years ago
12

Determine whether the function f(x) = 4x3 is even or odd.

Mathematics
2 answers:
nikitadnepr [17]3 years ago
5 0

Answer:

Its the last option.

Step-by-step explanation:

f(x) = 4x^3.

f(-x) =

4(-x)^3 = - 4x^3

so f(x) = = -f(x)  so the function is odd.

Annette [7]3 years ago
3 0

Answer:

its D

Step-by-step explanation:

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3 years ago
Can someone help me please I need help really bad. I'll give BRAINLIEST and give points
alukav5142 [94]

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5 0
3 years ago
A website manager has noticed that during the evening​ hours, about 5 people per minute check out from their shopping cart and m
Over [174]

Answer:

a) Poisson distribution

b) 99.33% probability that in any one minute at least one purchase is​ made

c) 0.05% probability that seven people make a purchase in the next four ​minutes

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.

5 people per minute check out from their shopping cart and make an online purchase.

This means that \mu = 5

a) What model might you suggest to model the number of purchases per​ minute? ​

The only information that we have is the mean number of an event(purchases) in a time interval. Each event is also independent fro each other. So you should suggest the Poisson distribution to model the number of purchases per​ minute.

b) What is the probability that in any one minute at least one purchase is​ made? ​

Either no purchases are made, or at least one is. The sum of the probabilities of these events is 1. So

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

We want to find P(X \geq 1)

So

P(X \geq 1) = 1 - P(X = 0)

In which

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

P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067

1 - 0.0067 = 0.9933.

99.33% probability that in any one minute at least one purchase is​ made

c) What is the probability that seven people make a purchase in the next four ​minutes?

The mean is 5 purchases in a minute. So, for 4 minutes

\mu = 4*5 = 20

We have to find P(X = 7).

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

P(X = 0) = \frac{e^{-20}*(20)^{7}}{(7)!} = 0.0005

0.05% probability that seven people make a purchase in the next four ​minutes

8 0
3 years ago
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