Solve for x show all your work. 10x + 15 = 5x

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 = the interval of time between the eruption

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 = 0.3264

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 $\bar X$ = sample time intervals between the eruption

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 = 0.0526

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 $\bar X$ = sample time intervals between the eruption

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 = 0.0222

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

What is the rate of change of y with respect to x for this function?

lilavasa [31]

Answer:

Given the function: y=f(x) = 3x+2

when x=-2 at the beginning of the interval [-2, 5],

then;

y = 3x+2 begins at

y= 3(-2)+2 = -6+2= -4.

and

when x=5 at the end of the interval [-2, 5],

y = 3x+2 ends up at

y= 3(5)+2 = 15+2= 17.

So,

y has changed -4 to 17, which is a change of 17-(-4)= 17+4 = 21

and x has changed from -2 to 5, which is a change of 5-(-2)=5+2=7

So, the average rate of change of y with respect to x over the interval

[-2, 5] is given by ;

=

Therefore, the average rate of change y with respect to x over the interval is, 3

Step-by-step explanation:

4 0

3 years ago

What is the approximation for the area of this circle? Use 3.14 to approximate pi. 12.6m 25.1m 50.2m 158.0m

Semenov [28]

Answer:

You times pi by the diameter

first set of answers are

1. 5.1

2 56.5

3 25.1

4 34.5

5 9.4

You times pi by r^2

second set of answers are

1 28.3

2 254.3

3 78.5

4 50.2

5 30.2

8 0

3 years ago

Read 2 more answers

I need to 2,4,6,8 But please send me the work too

Ugo [173]

ok for number 2.) 2+8+1+5+1+3+1+7+5+4+3+1=41 and 41 divided by the number of number which is 12= 3.41666666667

the mean for number 2 is 3.41666666667.

for the median you put the number from least to greatest, like so:

1, 1, 1, 1, 2, 3, 3, 4, 5, 5, 7, 8.

The middle is 3 and 3 so 3+3 divided by 2 is 6 so the median is 6.

I cant do mode but range is 8-1 divided by 2

4 0

3 years ago

What calender year does x=12 represent?

Mazyrski [523]

Answer:

December

Step-by-step explanation:

There are 12 months in a year and if the x value is equal to 12, and December is the 12th month of the year then x=December in the calender year.

5 0

3 years ago

Read 2 more answers

Solve for x show all your work. 10x + 15 = 5x - 10​

Solve for x show all your work. 10x + 15 = 5x - 10