In how many distinct ways can the

Nimfa-mama [501]

Answer:

Follow the instructions on the sheet. I believe this is an example question, so it doesn't need to be solved.

8 0

3 years ago

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 = the interval of time between the eruptions

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 = 0.3336

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 $\bar X$ = sample mean time between the eruptions

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 = 0.0582

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 $\bar X$ = sample mean time between the eruptions

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 = 0.0055

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

4 years ago

From delta math: Subtract -2x^2 + 3x - 9 from 8x^2 + 10x - 10

Leto [7]

Answer:

$10x^2 + 7x - 1$

Step-by-step explanation:

Step 1: Build the expression in numerical form.

$(8x^2 +10x - 10) - (-2x^2 + 3x - 9)$

Step 2: Distribute the negative sign.

$8x^2 +10x -10 + (-1)(-2x^2) + (-1)(3x) + (-1)(-9)$
$8x^2 + 10x - 10 + 2x^2 + (-3x) + 9$

Step 3: Combine like terms.

$(8x^2 +2x^2) + (10x -3x) + (-10+9)$
$10x^2 +7x -1$

Therefore, the answer is $10x^2 + 7x -1$ .

4 0

3 years ago

A salesperson is paid a flat rate plus a fixed percentage of her sales. Last month, she sold $16,000 worth of goods and was paid

Feliz [49]

Answer:

She will be paid $1,350

Step-by-step explanation:

Linear Modeling

Some events can be modeled as linear functions. If we are in a situation where a linear model is suitable, then we need two sample points to make the model and predict unknown behaviors.

The linear function can be expressed in the slope-intercept format:

y=mx+b, where m and b are constants.

The payments for a salesperson will be linearly modeled. There are two known points: When the sales were $16,000, the payment was $1,600. This makes the point (16,000;1,600).

We also know when the sales were $12,000, the payment was $1,400. The point is (12,000;1,400)

Let's use the points to find the values of m and b.

Using (16,000;1,600):

1,600=m*16,000+b

Using (12,000;1,400):

1,400=m*12,000+b

Subtracting both equations:

200=16,000m-12,000m

200=4,000m

Solving:

m=200/4,000=0.05

Using the first equation and the value of m:

1,600=0.05*16,000+b

1,600=800+b

Solving:

b=800

The equation is now complete:

y=0.05x+800

She sold $11,000 this month, so the payment is:

$y=0.05\cdot 11,000+800$

Y=550+800=1,350

She will be paid $1,350

3 0

3 years ago

a soccar team spent $55 dollars on supplies for a car wash then earned $275 whats their total after paying for supplies

IrinaK [193]

To find what profit they made, subtract what they earned from what they spent money on. 275-55=220. They made a total of $220 after paying for supplies. I hope this helps! :)

8 0

4 years ago