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Lyrx [107]
2 years ago
14

A website randomly creates an initial password for people when they first sign up for an account. The password consists of five

letters, and cannot include numbers or special characters. The letters of the password cannot repeat. What is the approximate probability that a password will have no vowels? What is the approximate probability that the first letter of the password will be m?
Mathematics
2 answers:
Law Incorporation [45]2 years ago
7 0
The answers are C: 0.30934 for the first one, and B: 0.03846 for the second. 
sashaice [31]2 years ago
3 0
This is a non replacement question, so each letter decreases by 1.

No vowels
The question actually starts right here. The preamble tells us that we can only use letters, no numbers, and no symbols. We are not asked to do anything. The first question makes no mention of there being anything special about m. It can be there or it most likely is not. This part of the problem is the first part.

First Part
I'm not going to deal with y. I will declare it to be a consonant.
There are 5 vowels. None of them can be chosen.
21 letters remain (including the m since it has not been limited in any way yet).

21 * 20 * 19 * 18 * 17
================
26 * 25 * 24 * 23 * 22

2441880 / 7893600
0.30934
Note: If your calculator has nPr on it, you can enter 21 nPr 5 to eliminate all the typing.

Second Part
No vowels and the first letter MUST be m
1 * 20 * 19 * 18 * 17/( 26 * 25 * 24 * 23 * 22)
116200/7893600 = 0.01473

Vowels and the first letter m.
1 * 25 * 24 * 23 * 22/ 7893600 = 0.03846

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The mean annual salary for intermediate level executives is about $74000 per year with a standard deviation of $2500. A random s
lidiya [134]

Answer:

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

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 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 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.

In this question, we have that:

\mu = 74000, \sigma = 2500, n = 36, s = \frac{2500}{\sqrt{36}} = 416.67

What is the probability that the mean annual salary of the sample is between $71000 and $73500?

This is the pvalue of Z when X = 73500 subtracted by the pvalue of Z when X = 71000. So

X = 73500

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

By the Central Limit Theorem

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

Z = \frac{73500 - 74000}{416.67}

Z = -1.2

Z = -1.2 has a pvalue of 0.1151

X = 71000

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

Z = \frac{71000 - 74000}{416.67}

Z = -7.2

Z = -7.2 has a pvalue of 0.

0.1151 - 0 = 0.1151

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

8 0
2 years ago
On a coordinate plane, a curved line begins at point (1, 2) and ends at (5, 4). What is the range of the function on the graph?
slavikrds [6]

Answer:

Step-by-step explanation:

2<y<4

all real numbers >2 and < 4

5 0
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X+y=425. This is as simple it can be, unless theres more information. x is the distance traveled on the first day and y is the distance traveled on the second day. 


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