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kolbaska11 [484]
3 years ago
6

Jocelyn estimates that a piece of wood measures 5.5 cm, if it actually measures 5.62 cm, what is the percent error of Jocelyns e

stimate
Mathematics
2 answers:
Paladinen [302]3 years ago
8 0
Percent error=error/correct times 100
error=5.62-5.5=0.12
correct=5.62

percent error=0.12/5.62 times 100=0.02135231316725978647686832740214 times 100=
2.135231316725978647686832740214%
Marysya12 [62]3 years ago
7 0
The percent error is -2.1352% of Jocelyn's estimate.
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The time a randomly selected individual waits for an elevator in an office building has a uniform distribution with a mean of 0.
Amiraneli [1.4K]

Answer:

The mean of the sampling distribution of means for SRS of size 50 is \mu = 0.5 and the standard deviation is s = 0.0409

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size, of at least 30, can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 0.5, \sigma = 0.289

What are the mean and standard deviation of the sampling distribution of means for SRS of size 50?

By the Central Limit Theorem

\mu = 0.5, s = \frac{0.289}{\sqrt{50}} = 0.0409

The mean of the sampling distribution of means for SRS of size 50 is \mu = 0.5 and the standard deviation is s = 0.0409

Does it matter that the underlying population distribution is not normal?

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

What is the probability a sample of 50 people will wait longer than 45 seconds for an elevator?

We have to use 45 seconds as minutes, since the mean and the standard deviation are in minutes.

Each minute has 60 seconds.

So 45 seconds is 45/60 = 0.75 min.

This probability is 1 subtracted by the pvalue of Z when X = 0.75. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.75 - 0.5}{0.0409}

Z = 6.11

Z = 6.11 has a pvalue of 1

1-1 = 0

0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.

8 0
3 years ago
Please answer ASAP I will give you five stars
Nataliya [291]

Known :

r = 6

h = 8

Asked :

V = ...?

Answer :

V = ⅓πr²h

= ⅓ × 3.14 × 6² × 8

= ⅓ × 3.14 × 36 × 8

= 3.14 × 12 × 8

= 301.44 units²

= <u>3</u><u>0</u><u>0</u><u> </u><u>units²</u> (rounded)

So, the volume of the cone is 300 units²

<em>Hope </em><em>it </em><em>helpful </em><em>and </em><em>useful </em><em>:</em><em>)</em>

3 0
3 years ago
you spin the spinner, flip a coin, then spin the spinner again. find the probability of the compound event. Spinning a 4, flippi
Tanya [424]
How many numbers are on the spinner? 
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3 years ago
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Translate to an algebraic expression: 15 less than six times a number, y.
Natalija [7]

Answer:

4

Step-by-step explanation:

6 0
3 years ago
Hello can anyone please help me with this assignment it’s due tomorrow please and thanks !
snow_lady [41]

Answer:

PART 2,4,6 are not solvable since they do not have any number

Part 1

1 = 61

2 = 119

3 = 61

4 = 119

5 = 61

6 = 119

7 = 61

8 = 119

Part 3

1 = 128

2 = 52

3 = 128

4= 52

5 = 128

6 =52

7 = 128

8 =52

Part 5

1 = 135

2 = 45

3 = 135

4 = 45

5= 135

6= 45

7=135

8= 45

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3 years ago
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