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Tasya [4]
2 years ago
14

Nikita ran a 5-kilometer race in 39 minutes (0.65 of an hour) without training beforehand. In the first part of the race, her av

erage speed was 8.75 kilometers per hour. For the second part of the race, she started to get tired, so her average speed dropped to 6 kilometers per hour. Which expression represents Nikita’s distance for the second part of the race?
Mathematics
1 answer:
s344n2d4d5 [400]2 years ago
6 0

Nikita's distance from her for the second part of the race was 1.5 kilometers.

Since Nikita ran a 5-kilometer race in 39 minutes (0.65 of an hour) without training beforehand, and in the first part of the race, her average speed was 8.75 kilometers per hour, while for the second part of the race, she started to get tired, so her average speed dropped to 6 kilometers per hour, to determine which expression represents Nikita's distance for the second part of the race the following calculation must be performed:

  • 8.75 x 0.65 + 6 x 0 = 5.68
  • 8.75 x 0.50 + 6 x 0.15 = 5.275
  • 8.75 x 0.40 + 6 x 0.25 = 5
  • 100 = 60
  • 40 = X
  • 40 x 60/100 = X
  • 24 = X
  • 39 - 24 = 15
  • 15 = 1/4 hour
  • 6/4 = 1.5

Therefore, Nikita's distance from her for the second part of the race was 1.5 kilometers.

Learn more about maths in brainly.com/question/22791791

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3 years ago
The average daily jail population in the United States is 706,242. If the distribution is normal and the standard deviation is 5
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a. The probability that on a randomly selected day, the jail population

is greater than 750,000 is 20.1%

b. The probability that on a randomly selected day, the jail population is

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Step-by-step explanation:

The given is:

1. The average daily jail population in the United States is 706,242

2. The distribution is normal and the standard deviation is 52,145

3. We need to find the probability that on a randomly selected day,

    the jail population is greater than 750,000

4. We need to find the probability that on a randomly selected day,

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a.

At first find z-score

∵ z = (x - μ)/σ, where x is the score, μ is the mean and σ is the standard

   deviation

∵  x = 750,000 , μ = 706,242 and σ = 52,145

∴ z = \frac{750,000-706,242}{52,145} ≅ 0.84

Use the normal distribution table of z to find the area to the right of

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- But we are interested in x > 750,000, we need the area to the

  right of z-score

∴ P(x > 750,000) = 1 - 0.79955 = 0.2005

∴ P(x > 750,000) = 0.2005 × 100% = 20.1%

The probability that on a randomly selected day, the jail population is

greater than 750,000 is 20.1%

b.

We will find z-score for 600,000 < x < 700,000

∵ z = \frac{600,000-706,242}{52,145} ≅ -2.04

∵ z = \frac{700,000-706,242}{52,145} ≅ -0.12

Use the normal distribution table of z to find the area between

the two z-values

∵ The corresponding area to z-score of -2.04 is 0.02068

∵ The corresponding area to z-score of -0.12 is 0.45224

- To find P(600,000 < x < 700,000) subtract the two values above

∴ P(600,000 < x < 700,000) = 0.45224 - 0.02068 = 0.4316

∴ P(600,000 < x < 700,000) = 0.4316 × 100% = 43.2%

The probability that on a randomly selected day, the jail population is

between 600,000 and 700,000 is 43.2%

Learn more:

You can learn more about z-score in brainly.com/question/7207785

#LearnwithBrainly

5 0
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Answer:

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