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FromTheMoon [43]
3 years ago
8

NEED HELP ASAP Solve the equation or inequality for the unknown number. Show your work.

Mathematics
1 answer:
amid [387]3 years ago
6 0

Answer:

5

Step-by-step explanation:

3(14+x) = 57

42 +3x = 57

3x = 15

x = 5

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Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

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

So, X ~ Normal(\mu=75, \sigma^{2} =20)

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 between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</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 between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</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 between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
3 years ago
when (2x - 3)^2 is subtracted from 5x^2, what is the result? And how can I show the answer? Thank you x
Leni [432]
5x^2 - (2x - 3)^2 = 
5x^2 - ((2x - 3)(2x - 3)) =
5x^2 - (4x^2 - 6x - 6x + 9) =
5x^2 - (4x^2 - 12x + 9) =
5x^2 - 4x^2 + 12x - 9 =
x^2 + 12x - 9 <===
3 0
3 years ago
Read 2 more answers
HELP FAST I WILL MARK AS BRAINLIEST.
vodka [1.7K]

Answer:

  • s = 1/3(a + b + c + 16)

Step-by-step explanation:

<h3>Given</h3>
  • If (s -a) = 5cm, (s -b) = 10cm, (s -c) = 1cm
  • s = ?
<h3>Solution</h3>

<u>Add up the 3 equations and solve for s:</u>

  • s - a + s - b + s - c = 5 + 10 + 1
  • 3s = 16 + a + b + c
  • s = 1/3(a + b + c + 16)
7 0
3 years ago
Hector had y dollars in his savings account. He made a deposit of twenty-dollar bills and dollar coins. He had four times as man
Mazyrski [523]

Answer:

Y= $240

Step-by-step explanation:

Since he had 60 dollar bills and the dollar coins were 4 times that amount, you just multiply 60 by 4.

5 0
3 years ago
9n - 7 = 5n - 19<br> What is n and why did you do that
tamaranim1 [39]

14n=-26

i tried at least

5 0
3 years ago
Read 2 more answers
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