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Troyanec [42]
3 years ago
8

Andre looks at a box of paper clips. He says: “I think the number of paper clips in the box is less than 1,000.” Lin also looks

at the box. She says: “I think the number of paper clips in the box is more than 500.”
Write an inequality to show Andre's statement, using p for the number of paper clips.
Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
7 0

Answer:

<em>500 </em>\leq<em> </em><em>p </em>\leq<em> </em><em>1000</em>

Step-by-step explanation:

Since Andre if giving a rough estimate about how much paper clips are in the box, we will use the \leq (less than or equal to)

Your X value will be your actual number, the equation shows us that p is anywhere between 500 - 1000

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A ferris wheel goes around once every 20 seconds . How many times will a rider be ag thetop during an 8-minuteride?
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Answer:

The answer is B.

Step-by-step explanation:

There are three 20 second time periods in 1 minute. Three times eight is equal to 24.  

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There are 5people trying for 3openings how many possible outcomes are there ​
Zigmanuir [339]

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60

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Alternately,

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Each day, Robin commutes to work by bike with probability 0.4 and by walking with probability 0.6. When biking to work injuries
kvasek [131]

Answer:

64.65% probability of at least one injury commuting to work in the next 20 years

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)!}&#10;

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Each day:

Bikes to work with probability 0.4.

If he bikes to work, 0.1 injuries per year.

Walks to work with probability 0.6.

If he walks to work, 0.02 injuries per year.

20 years.

So

\mu = 20*(0.4*0.1 + 0.6*0.02) = 1.04

Either he suffers no injuries, or he suffer at least one injury. The sum of the probabilities of these events is decimal 1. So

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

We want P(X \geq 1). Then

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

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}&#10;

P(X = 0) = \frac{e^{-1.04}*1.04^{0}}{(0)!} = 0.3535&#10;

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.3535 = 0.6465

64.65% probability of at least one injury commuting to work in the next 20 years

3 0
3 years ago
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