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r-ruslan [8.4K]

3 years ago

7

The marine life store would like to set up fish tanks that contain equal numbers of angel fish , sword tales , and guppies. what

is the greatest number of tanks that can be set up if the store has 12 angel fish, 24 sword tales and 30 guppies?

1 answer:

Shalnov [3]3 years ago

7 0

How do people......I just don't understand this

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Read 2 more answers

A subway has good service 70% of the time and runs less frequently 30% of the time because of signal problems. When there are si

nordsb [41]

Answer:

Step-by-step explanation:

Let GS denote the good service and SP denote the signal problem.

A subway has good service 70% of the time, that is, $P(GS)=0.7$ and a subway runs less frequently 30% of the time because of the signal problems, that is, $P(SP)=0.3$ .

If there are signal problems, the amount of time T in minutes that have to wait at the platform is described by the probability density function given below:

$P_{T|SP}(t)=0.1e^{0.1t}$

If there is good service, the amount of time T in minutes that have to wait at the platform is described the probability density function given below:

$P_{T|GOOD}(t)=0.3e^{0.3t}$

(a)

The probability that you wait at least 1 minute if there is good service P(T ≥ 1| GS) is obtained as follows:

$P(T\geq 1|GS)=\int\limits^{\infty}_1 {0.3e^{-0.3t}} \, dt\\\\=0.3\int\limits^{\infty}_1 {e^{-0.3t}} \, dt\\\\=0.3[(\frac{e^{-0.3t}}{-0.3})]\\\\=-(e^{-0.3t})\limits^{\infty}_1\\\\=-(0-e^{-0.3})\\=0.74$

(b)

The probability that you wait at least 1 minute if there is signal problems P(T ≥ 1| SP) is obtained as follows:

$P(T\geq 1|SP)=\int\limits^{\infty}_1 {0.1e^{-0.1t}} \, dt\\\\=0.1\int\limits^{\infty}_1 {e^{-0.1t}} \, dt\\\\=0.1[(\frac{e^{-0.1t}}{-0.3})]\\\\=-(e^{-0.1t})\limits^{\infty}_1\\\\=-(e^{\infty}-e^{-0.1})\\=-(0-0.904)\\=0.904$

(c)

After 1 minute of waiting on the platform, the train is having signal problems follows an

exponential distribution with parameter $\lambda= 0.1$

The probability that the train is having signal problems based on the fact that will be at least 1 minute long is obtained using the result given below:

$P(SP|T\geq 1)=\frac{P(T\geq 1|SP)P(SP)}{P(T\geq 1)}$

$P(T\geq 1|GS)=0.74, P(T\geq 1|SP)=0.904$

Now calculate the $P(T \geq 1)$ as follows:

$P(T \geq 1)=P(T\geq 1|SP)P(SP)+P(T\geq 1|GS)P(GS)\\=(0.904)(0.3)+(0.74)(0.7)=0.7892$

The probability that the train is having signal problems based on the fact that will be at least 1 minute long is calculated as follows:

$P(SP|T\geq 1)= \frac{0.904 \times 0.3}{0.7892}
= 0.3436
$

Hence, the probability that the train is having signal problems based on the fact that will be at least 1 minute long is $0.3436$ .

(d)

After 5 minutes of waiting on the platform, the train is having signal problems follows an exponential distribution with parameter $\lambda= 0.1$ .

The probability that the train is having signal problems based on the fact that will be at least 5 minutes long is obtained using the result given below:

$P(SP|T\geq 5)=\frac{P(T\geq 5|SP)P(SP)}{P(T\geq 5)}$

First, calculate the $P(T\geq 5|SP)$ as follows:

$P(T\geq 5|SP)=\int\limits^{\infty}_5 {0.1e^{-0.1t}} \, dt\\\\=0.1\int\limits^{\infty}_5 {e^{-0.1t}} \, dt\\\\=0.1[(\frac{e^{-0.1t}}{-0.1})]\\\\=-(e^{-0.1t})\limits^{\infty}_5\\\\=-(e^{\infty}-e^{-0.5})\\=-(0-0.6065)\\=0.6065$

Now, calculate the $P (T\geq5|GS )$ as follows:

$P(T\geq 5|GS)=\int\limits^{\infty}_5 {0.3e^{-0.3t}} \, dt\\\\=0.3\int\limits^{\infty}_5 {e^{-0.3t}} \, dt\\\\=0.3[(\frac{e^{-0.3t}}{-0.3})]\\\\=-(e^{-0.3t})\limits^{\infty}_5\\\\=-(0-e^{-1.5})\\=0.2231$

Now, calculate the $P (T \geq 5)$ as follows:

$P(T \geq 5)=P(T\geq 5|SP)P(SP)+P(T\geq 5|GS)P(GS)\\=(0.6065)(0.3)+(0.2231)(0.7)=0.3381$

The probability that the train is having signal problems based on the fact that will be at least 5 minutes long is calculated as follows:

$P(SP|T\geq 5)= \frac{0.6065 \times 0.3}{0.3381}
= 0.5381
$

Hence, the probability that the train is having signal problems based on the fact that will be at least 1 minute long is $0.5381$ .

6 0

2 years ago

Determine if the statement is true or false. A linear system with three equations and two variables must be inconsistent. True F

Arturiano [62]

Answer:

yes it is true that a linear system of three equations with two variables is inconsistent because when we solve the equations variables can be obtained by only first two equations which are considered first. and third equation is useless.

6 0

3 years ago