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Lemur [1.5K]
3 years ago
9

Find the average rate of change of the given function on the interval [1,6]

Mathematics
1 answer:
AleksAgata [21]3 years ago
3 0
It is fifteen which isA
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TIME REMAINING
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Answer:

<h2>YOUR CORRECT ANSWER IS(B.)</h2>

Step-by-step explanation:

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A locker requires a three-digit code to open the lock. The code must contain one letter and two numbers, and no letter or number
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not sure what the dots are supposed to mean..

but i think it's 6B5??

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3 years ago
How do you tell time on a clock
dimaraw [331]

Answer:

Each of those dashes are 5 minute for the small hand

and each of those dashes are 1 minutes for the long hand

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Please help &amp; explain PLEASE
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3 years ago
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In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t
Angelina_Jolie [31]

Answer:

P(X = 0) = 0.027

P(X = 1) = 0.189

P(X = 2) = 0.441

P(X = 3)= 0.343

Step-by-step explanation:

The probability mass function P(X = x) is the probability that X happens x times.

When n trials happen, for each x \leq n, the probability mass function is given by:

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

In which p is the probability that the event happens.

pe_{n,x}, is the permutation of n elements with x repetitions(when there are multiple events happening(like one passes and two not passing)). It can be calculated by the following formula:

pe_{n,x} = \frac{n!}{x!}

The sum of all P(X=x) must be 1.

In this problem

We have 3 trials, so n = 3

The probability that a wafer pass a test is 0.7, so p = 0.7

Determine the probability mass function of the number of wafers from a lot that pass the test.

P(X = 0) = (0.7)^{0}.(0.3)^{3} = 0.027

P(X = 1) = pe_{3,1}.(0.7).(0.3)^{2} = 0.189

P(X = 2) = pe_{3,2}.(0.7)^{2}.(0.3) = 0.441

P(X = 3) = (0.7)^{3} = 0.343

5 0
3 years ago
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