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3 years ago

When multiplying an inequality by a negative number, make sure to flip the inequality sign.

Mathematics

1 answer:

timama [110]3 years ago

3 0

Answer:

the answer is false but when it is divided by a negative number then you flip the inequality sign

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A software company is releasing one of their products on a CD. The manufacturer charges a $5,500 setup fee and $1.00 per CD. App

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Given:
set-up fee 5,500
variable fee 1 per CD.

Cost per CD is $5.50

5.50 - 1= 4.50

5,500 / 4.50 = 1,222 CDs Choice A.

1*1,222 = 1,222
1,222 + 5,500 = 6,722
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3 years ago

This person has a 75% chance of a full recovery. Is this classical probability, empirical or subjective

RideAnS [48]

Answer:

Subjective probability

Step-by-step explanation:

At first we should know the following:

Classical probability ⇒ when there are n equally likely outcomes.

Subjective probability ⇒ is based on whatever information is available.

Empirical probability ⇒ when the number of times the event happens is divided by the number of observations.

So, according to the previous definitions:

This person has a 75% chance of a full recovery

There is no equally likely outcomes, and the percentage of full recovery is based on the information available about the person and also it is based on educated guess.

So, this is Subjective probability

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3 years ago

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3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

marshall27 [118]

Answer:

1. 0.0000454

2. 0.01034

3. 0.0821

4. 0.918

Step-by-step explanation:

Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,

$X \sim Poi(\lambda = 10)$

1. Requires us to compute

$P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454$

2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).

$P(Y=0) + P(Y \geq 1) = 1 \implies P(Y \geq 1) = 1 -P(Y=0) = 1 - 0.0821 = 0.918$

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3 years ago