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Vikentia [17]
3 years ago
14

PLEASE HELPPPPPPPPP!!!!?????

Mathematics
1 answer:
natta225 [31]3 years ago
8 0

Answers:

  • Mean = 69
  • Median = 69
  • Mode = {68, 70}

===============================================

Explanations:

To get the mean, you add up all of the values and then you divide by N, where N is the number of values in the list. In this case, you'll divide by N = 8.

Adding up all the values gets you 70+68+70+72+68+66+65+73 = 552. Dividing that over 8 then leads to the arithmetic mean of 552/8 = 69

-------

To find the median, you'll first sort the values from smallest to largest. You should have this set {65, 66, 68, 68, 70, 70, 72, 73} after sorting. Next, cross off the smallest and largest values to end up with this smaller set {66, 68, 68, 70, 70, 72}. We do this so we can locate the middle-most item.

Repeat the process of crossing off the smallest and largest items to get {68, 68, 70, 70}. Do so one more time and we get {68,70}.

The midpoint of those two values is (68+70)/2 = 138/2 = 69. The median is 69. The mean and median may not always be the same value.

--------

The mode is the most frequent value. In this case, we have 68 and 70 show up exactly twice for each unique value. This is the most compared to the other unique values that only show up exactly once.

It is possible to have more than one mode, as long those values are more frequent than the rest. We denote the mode as a set, so the mode is {68, 70}

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Answer:

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Step-by-step explanation:

What is the percent decrease if the original price is $582 and it went down 30%

or decreased 30%

Check:

582 x 0.70 = 407.4

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3 years ago
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Answer:

The probability is 0.64

Step-by-step explanation:

What we want to calculate here is conditional probability.

Let P(A )= Probability of using route A = 50% = 0.5

Let P( B )= probability of using route B = 25% = 0.25

Let P (C )= probability of using route C = 25% = 0.25

Let T be the probability that he will be in a traffic Jam

The probability that he will be in a traffic Jam if he uses route A = 80%

Mathematically this is written as P( T | A) which is read as probability of T given A

so P( T | A) = 0.8

Same way for B and C which can be written as follows;

P( T | B) = 60% = 0.6

P( T | C) = 30% = 0.3

Now, what do we want to calculate?

He is in a traffic Jam, and we want to find the probability that he used route A. This means we want to find P(A) given T which can be written mathematically as P ( A | T)

We can find this using the other parameters and especially the equation below;

P ( A | T) = P(A) • P( T| A) / {P(A) • P ( T | A) + P(B)• P(T| B) + P(C) • P(T|C)

P( A | T) = (0.5 * 0.8)/ ( 0.5)(0.8) + (0.25)(0.6) + (0.25)(3) = 0.4/(0.4 + 0.75 + 0.075) = 0.4/0.625 = 0.64

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Which equation is represented by the graph below?
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The correct answer is the first choice
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David's score on his fifth test would be an 87. If you add up all the scores from his first to his fourth test, it would 363. However, since there are five tests, in order to get a 90 percent, the sum of all five test scores would be 450. 450 - 363 = 87.
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3 years ago
The Office of Student Services at a large western state university maintains information on the study habits of its full-time st
Vera_Pavlovna [14]

Answer:

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 20 hours, standard deviation of 6:

This means that \mu = 20, \sigma = 6

Sample of 150:

This means that n = 150, s = \frac{6}{\sqrt{150}}

What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours?

This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.5. So

X = 21

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

By the Central Limit Theorem

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

Z = \frac{21 - 20}{\frac{6}{\sqrt{150}}}

Z = 2.04

Z = 2.04 has a p-value of 0.9793

X = 19.5

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

Z = \frac{19.5 - 20}{\frac{6}{\sqrt{150}}}

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Z = -1.02 has a p-value of 0.1539

0.9793 - 0.1539 = 0.8254

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

3 0
3 years ago
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