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marshall27 [118]
3 years ago
13

Fission tracks are trails found in uranium-bearing minerals, left by fragments released during fission events. An article report

s that fifteen tracks on one rock specimen had an average track length of 12 μm with a standard deviation of 2 μm. Assuming this to be a random sample from an approximately normal population, find a 99% confidence interval for the mean track length for this rock specimen. Round the answers to three decimal places.
Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
4 0

Answer:

Mean track length for this rock specimen is between 10.463 and 13.537

Step-by-step explanation:

99% confidence interval for the mean track length for rock specimen can be calculated using the formula:

M±\frac{t*s}{\sqrt{N}} where

  • M is the average track length (12 μm) in the report
  • t is the two tailed t-score in 99% confidence interval (2.977)
  • s is the standard deviation of track lengths in the report (2 μm)
  • N is the total number of tracks (15)

putting these numbers in the formula, we get confidence interval in 99% confidence as:

12±\frac{2.977*2}{\sqrt{15}} =12±1.537

Therefore, mean track length for this rock specimen is between 10.463 and 13.537

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luda_lava [24]

Answer:

C

Step-by-step explanation:

A regression model is a function which can be used to predict behavior of the model. It takes data points in and forms a general pattern for the pattern using an equation. This model y = 12.3 + 0.12x is a linear regression model which means it has linear behavior (a line) where x is total yards gained and y is number of points scored. If one additional yard is gained then the score is likely to be 12.3 + 0.12(1) = 12.42. If two additional yards is gained then the score is likely to be 12.3 + 0.12(2) = 12.54. Notice the score increases by 0.12 points for every yard gained. This is answer choice C.

a.)The average points scored for teams who gain zero yards during a game is -12.3 points.

b.) The average yards gained will increase by .12 for every additional point scored.

c.) The average change in points scored for each increase of one yard will be 0.12.

d.) The average number of points scored per game is 12.3.

8 0
3 years ago
Jennifer and Kamlee are walking to school. After 20 minutes, Jennifer has walked 4/5 miles. After 25 minutes, Kamlee has walked
Nimfa-mama [501]

Answer:

Jennifer's speed is 2.4 miles/hour

Kamlee's speed is 2 miles/hour

Step-by-step explanation:

To find their speed,

Speed can be determined from the formula below,

Speed = \frac{Distance}{Time}

Also, to determine the speed in miles per hour (miles/hour), we will convert the time spent by each of them to hour (NOTE: 60 minutes = 1 hour)

For Jennifer,

Distance = 4/5 miles

Time = 20 mins (Convert to hour)

∴Time = (20/60) hour = 1/3 hour

Hence,

Speed = \frac{Distance}{Time}

Speed = \frac{4/5  miles}{1/3hour}

Speed = \frac{4}{5} \times \frac{3}{1}

Speed = \frac{12}{5}

Speed = 2.4 miles/hour

Hence, Jennifer's speed is 2.4 miles/hour

For Kamlee,

Distance = 5/6 miles

Time = 25 minutes (Convert to hour)

Time = (25/60) hour = 5/12 hour

Hence,

Speed = \frac{Distance}{Time}

Speed = \frac{5/6miles}{5/12 hour}

Speed = \frac{5}{6} \times \frac{12}{5}

Speed = 2 miles/hour

Hence, Kamlee's speed is 2 miles/hour

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3 years ago
Evaluate 4 ^ x for (a) x = - 2 and (b) x = 3 . Write your answers in simplest form
devlian [24]

Answer:

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Step-by-step explanation:

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3 years ago
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Length of the radius = ½ of the distance between the two endpoints

Distance between (-2, 4) and (3, 1) using the formula, d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

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(3, 1) = (x_2, y_2)

d = \sqrt{(3 - (-2))^2 + (1 - 4)^2}

d = \sqrt{(5)^2 + (-3)^2}

d = \sqrt{25 + 9} = \sqrt{34}

d = 5.8

Length of radius = ½ of 5.8 = 2.9 units.

5 0
3 years ago
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