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3 years ago

What is the maximum number of relative extrema contained in the graph of this function?

Mathematics

2 answers:

svp [43]3 years ago

8 0

Answer:

Step-by-step explanation:

Savatey [412]3 years ago

6 0

Answer:

the graph of this function has no local extrema

Step-by-step explanation:

A 5th-degree polynomial graph will have a maximum of 4 local extrema, in general. However, this function has none. The derivative of this function is positive everywhere, so there are no extrema.

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For question 6 pls help show work step by step

Alexxx [7]

The steps are labeled, and the chart on the side is to help you know which numbers are x, and which ones are y.

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4 years ago

What is the measure of 165 15 82.5 97.5

Zina [86]

Answer:

15°

Step-by-step explanation:

L MLP = 180-165=15°

_____

8 0

3 years ago

If a person runs 1/3 mile in 1/9 of an hour what is the unit rate per hour

Natasha2012 [34]

Answer:

3 miles per hour

Step-by-step explanation:

he runs 1/3 mile in 1/9 of an hour

then

he runs (we multiply both sides by 9)

9×(1/3) mile in 9×(1/9) of an hour

then

he runs

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then

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4 0

2 years ago

Assume that the number of messages input to a communication channel in an Exponential distribution with 7 messages arriving in a

NARA [144]

The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

According to the statement

we have given that the an Exponential distribution with 7 messages arriving in a 10 second period and we have to find the probability that more than 3 messages will arrive during a 30-second interval.

So, For this purpose, we know that the

The probability is the measure of the likelihood of an event to happen. It measures the certainty of the event.

And the given information is that :

3 messages will arrive during a 30-second interval.

Then

Probability = P(X=1) + P(X=2) + P(X=3).Then

The probability become according to the exponential distribution:

$P(X=1)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\\P(X=2)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\\P(X=3)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

And then substitute the values in it then

$Probability = \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\ +\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\ + \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

This is the probability.

So, The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

Learn more about probability here

brainly.com/question/24756209

#SPJ4

7 0

2 years ago

How do you fins the sum of 2/10 and20/100?

Olegator [25]

2/10+20/100
their lcd is 100
20/100+20/100
=40/100 or 2/4

3 0

3 years ago

Read 2 more answers