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patriot [66]
3 years ago
14

How do you solve number 8

Mathematics
2 answers:
wel3 years ago
8 0
2n+6=2n+6

0=0

inconsistent solutions
gizmo_the_mogwai [7]3 years ago
6 0
4n+6-2n=2(n+3)
2n+6=2n+6
(subtract 6 from both sides)
2n=2n
(subtract 2n from both sides)
0=0
Your answer is Infinite Solution
Hope this helps love! :)  
You might be interested in
Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis
nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = <u><em>the interval of time between the eruptions</em></u>

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( \frac{X-\mu}{\sigma} > \frac{82-72}{23} ) = P(Z > 0.43) = 1 - P(Z \leq 0.43)

                                                           = 1 - 0.6664 = <u>0.3336</u>

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

                                                           = 1 - 0.9418 = <u>0.0582</u>

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

                                                           = 1 - 0.9945 = <u>0.0055</u>

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

6 0
3 years ago
Q(a+1) - 2q(a) if q(x) = x²+3x+4​
siniylev [52]

q(x) =  {x}^{2} + 3x + 4

_________________________________

Step(1)

To find q(a) we just need to put a instead of x in q(x) function.

Let's do it...

q(a) =  {a}^{2} + 3a + 4

Multiply sides by -2 :

- 2q(a) =  - 2( {a}^{2} + 3a + 4)

- 2q(a) =  - 2 {a}^{2} - 6a - 8

_________________________________

Step (2)

To find q(a+1) we just need to put a+1 instead of x in q(x) function.

Let's do it...

q(a + 1) =  ({a + 1})^{2} + 3(a + 1) + 4 \\

q(a + 1) =  {a}^{2} + 2a + 1 + 3a + 3 + 4 \\

q(a + 1) =  {a}^{2} + 5a + 8

_________________________________

Step (3)

q(a + 1) - 2q(a) =

{a}^{2} + 5a + 8 - 2 {a}^{2} - 6a - 8 =  \\

-  {a}^{2} - a  =  - a(a + 1)

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

8 0
3 years ago
Find the<br> area of the figure.
EastWind [94]

Answer:

22 sq ft

Step-by-step explanation:

8 0
2 years ago
Find the area of Region IV for this figure. A) 16 square units B) 64 square units C) 32 square units D) 36 square units
mamaluj [8]
The answer is D 8 x 2
4 0
2 years ago
Let x = 6.
Mars2501 [29]

Put the value of x = 6 to the eqpression 5x² + x - 7:

5(6)² + 6 - 7 = 5(36) + 6 - 7 = 180 + 6 - 7 = 179

<h3>Answer: 179</h3>
5 0
3 years ago
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