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amm1812
3 years ago
10

120 pianists compete in a piano competition. (a) In the first round, 30 of the 120 are selected to go on to the next round. How

many different outcomes are there for the first round? (b) In the second round, the judges select the first, second, third, fourth and fifth place winners of the competition from among the 30 pianists who advanced to the second round. How many outcomes are there for the second round of the competition?
Mathematics
1 answer:
MissTica3 years ago
8 0

Answer:

Step-by-step explanation:

Given

There are 120 Pianist in a competition

(a)In first round 30 pianist is selected

this can be done in ^{120}C_{30} ways

=\frac{120!}{(120-30)!30!}

=\frac{120!}{90!30!}

(b)In the second round 5 pianist is selected out of 30

this can be doe by =\frac{30!}{(30-5)!5!}

=\frac{30!}{25!5!}

=142,506

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3 years ago
Estimate the value of the function at x = 2. Can someone help me with this?
Zielflug [23.3K]

Answer:

y ≈ 2.1

Step-by-step explanation:

If you look at the graph and locate on the x axis 2, you can see that directly above on the line is a little more than 2.

7 0
3 years ago
Given: g(x) = 3x² + 2 and h(x) = 2x + 2<br> Find: g(h(x))
kompoz [17]

Step-by-step explanation:

g(x) = 3x² + 2

h(x) = 2x + 2

g(h(x)) = g([2x + 2]) = 3[2x+2]² + 2

= 3[4x² + 8x + 4] + 2

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7 0
3 years ago
The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD pla
castortr0y [4]

Using the normal distribution, we have that:

a) The sketch of the situation is given at the end of this answer.

b) The probability is:

P(2.8 \leq X \leq 7) = 0.8284

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.

In this problem:

  • Mean of 4.1 years, thus \mu = 4.1.
  • Standard deviation of 1.3 years, thus \sigma = 1.3.

Item a:

The part between 2.8 and 7 years is shaded on the sketch given at the end of this answer.

Item b:

The probability is the <u>p-value of Z when X = 7 subtracted by the p-value of Z when X = 2.8</u>, thus:

X = 7:

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

Z = \frac{7 - 4.1}{1.3}

Z = 2.23

Z = 2.23 has a p-value of 0.9871.

X = 2.8:

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

Z = \frac{2.8 - 4.1}{1.3}

Z = -1

Z = -1 has a p-value of 0.1587.

0.9871 - 0.1587 = 0.8284, thus:

P(2.8 \leq X \leq 7) = 0.8284

A similar problem is given at brainly.com/question/25151638

7 0
3 years ago
A frequency bar graph is shown for the shoe size of a class of 9th graders at Brookwood High School. What is the probability tha
Lapatulllka [165]
I saw the image. I'll just write the values but I also attached the image needed for this problem.

shoe size     frequency
7                            7
8                           16 
9                           19
10                         27
11                         33 
12                         35
total                   137

Probability that a student wears size 12 shoes is 0.255 or 25.5%.

P = 35/137 = 0.255



7 0
4 years ago
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