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lesya [120]
3 years ago
6

Im stuck on this question, plz help

Mathematics
1 answer:
denis-greek [22]3 years ago
3 0

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

40 metres covered in 6 seconds

Max speed attained after 6seconds

75 meters covered after 11 seconds

Average Speed for first 40 meters :

Speed = distance / time

Speed = 40m / 6s

Speed = 6.67m/s

To obtain the maximum speed :

Next (75 - 40) meters = 35 meters was covered in (11 - 6)seconds = 5 seconds

Speed at this point is maximum :

Hence, maximum speed = (35m / 5s) = 7m/s

Suppose, Manuel runs for an additional z seconds after reaching max speed :

Distance from starting line 6+z seconds after race started?

Distance after 6 seconds = 40 metres

Distance after z seconds = 7 * z

Total distance = (40 + 7z)

What is Manuel's distance from the starting line x seconds after the race started (provided x≥6x)?

Distance for first 6 seconds = 40 meters + distance covered after 6 seconds = (7 * (x-6))

40 + 7(x - 6)

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b) 0.0042495

c) 0.048021

Hence, the probability of the mean weight that the 10 apples exceeds 8.2 ounces of this customer wil be higher than the last customer in part b because the number of random samples in part b is greater that that in part c

Step-by-step explanation:

mean of 7.5 ounces and a standard deviation of 1.33 ounce

When random number of samples are given, we solve using z score with the formula

= z = (x-μ)/σ/√n, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

n is the random number of samples

(a) One customer comes to this grocery store and randomly picks 25 apples. What is the mean and standard deviation of the distribution of the mean weight for 25 apples?

(b) What is the probability that the mean weight of 25 apples exceeds 8.2 ounces?

x > 8.2 ounces

= z = (x - μ)/σ/√n

z = (8.2 - 7.5)/1.33/√25

z = (8.2 - 7.5)/1.33/5

z = 2.63158

Probability value from Z-Table:

P(x<8.2) = 0.99575

P(x>8.2) = 1 - P(x<8.2) = 0.0042495

(c) Another customer randomly picks 10 apples. Will the probability of the mean weight that the 10 apples exceeds 8.2 ounces of this customer be higher than the last customer in part b?

For n= 10

x > 8.2 ounces

= z = (x - μ)/σ/√n

z = (8.2 - 7.5)/1.33/√10

z = 1.66436

Probability value from Z-Table:

P(x<8.2) = 0.95198

P(x>8.2) = 1 - P(x<8.2) = 0.048021

Hence, the probability of the mean weight that the 10 apples exceeds 8.2 ounces of this customer wil be higher than the last customer in part b because the number of random samples in part b is greater that that in part c

4 0
3 years ago
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