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CaHeK987 [17]
2 years ago
15

I need the right answer!thanks in advance^^ ​

Mathematics
1 answer:
LiRa [457]2 years ago
3 0

The questions are illustrations of combination and permutation.

<h3>The different arrangement of LOVE</h3>

From the question, we understand that the letter V must come first.

So, we have:

  • First letter = 1; i.e. V
  • The second letter can be any of the remaining 3
  • The third letter can be any of the remaining 2
  • The last letter can be the remaining 1

So, we have:

Arrangement = 1 * 3 * 2 * 1

Arrangement = 6

Hence, there are 6 different arrangement of the letters.

<h3>Ways to pick two vegetables</h3>

We have:

  • Vegetables, n = 5
  • To pick, r = 2

The number of different combination is:

Combination = ^5C_2

Evaluate

Combination = 10

Hence, there are 10 different ways

<h3>Number of 4-digit code</h3>

The numbers are given as {1,4,5,7,8}

So, we have:

  • The first digit can be any of the 5 digits
  • The second digit can be any of the remaining 4 digits
  • The third digit can be any of the remaining 3 digits
  • The fourth digit can be any of the remaining 2 digits

So, we have:

Digits = 5 * 4 * 3 * 2

Digits = 120

Hence, there are 120 different codes of 4-digit

<h3>Arrangement of books in a shelf</h3>

We have:

  • Books, n = 4

The number of arrangement is:

Arrangement = 4!

Evaluate

Arrangement = 24

Hence, there are 24 different arrangements

<h3>Outcomes in tossing a coin</h3>

A coin has 2 faces

When four coins are tossed, the number of outcomes is:

Outcome = 2 * 2 * 2* 2

Evaluate

Outcome = 16

Hence, there are 16 different outcomes

<h3>Ways of sitting in a row</h3>

The number of people is given as:

People = 5

So, we have:

  • The first person can sit on any of the 5 seats
  • The second person can sit on any of the 4 seats
  • The third person can sit on any of the 3 seats
  • The fourth person can sit on any of the 2 seats
  • The last person would sit on the remaining seat1

So, we have:

Ways = 5 * 4 * 3 * 2 * 1

Ways = 120

Hence, there are 120 different ways

Read more about combination and permutation at:

brainly.com/question/11732255

#SPJ1

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Answer:

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The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.
salantis [7]

a) X

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

\mu=78.1 (mean)

\sigma = 10.8 (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

X

And the probability for this to occur can be written as:

p(X

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between z=-\infty and z=Z, where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

p(X

And by looking at the z-score tables, we find that this probability is:

p(z

And so,

p(X

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

p(X>89.1)

First of all, we have to calculate the z-score corresponding to this value of X, which is:

Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

p(z>1.02) =p(z

Because the normal distribution is symmetric.

But from part b) we know that

p(z

Therefore:

p(X>89.1)=p(z>1.02)=0.1539

d)

Here we want to find the probability that the randomly chosen score is between 67.1 and 89.1, which can be written as

p(67.1

Or also as

p(67.1

Since the overall probability under the whole distribution must be 1.

From part b) and c) we know that:

p(X

p(X>89.1)=0.1539

Therefore, here we find immediately than:

p(67.1

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<span>data: 2, 5, 9, 11, 17, 19
</span>
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mean = average
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