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pickupchik [31]
3 years ago
8

A professor has 24 pairs of socks in his drawer. 18 pairs are black and the rest are brown. He always wears black pants. He does

not pay much attention to which socks he puts on each day. Supposed he selects one pair of socks on Monday and one on Tuesday. Since he has lots of clean socks in the drawer he does not do laundry. What is the likelihood that the professor's socks matched his pants on both days (i.e. he picked black socks)
Mathematics
1 answer:
lesya692 [45]3 years ago
4 0

Answer:

The likelihood that the professor's socks matched his pants on both days is 0.5625

Step-by-step explanation:

Total no. of pair of socks = 24

No. of black pairs = 18

We are given that  he selects one pair of socks on Monday and one on Tuesday. Since he has lots of clean socks in the drawer he does not do laundry.

So, It is  a case of replacement

He always wears black pants.

We are supposed to find he likelihood that the professor's socks matched his pants on both days i.e. Black socks both days.

So ,Probability of wearing black socks on Monday = \frac{18}{24}

So ,Probability of wearing black socks on Tuesday =\frac{18}{24}

So,the likelihood that the professor's socks matched his pants on both days = \frac{18}{24} \times \frac{18}{24}=0.5625

Hence the likelihood that the professor's socks matched his pants on both days is 0.5625

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pishuonlain [190]

The five-number summary and the interquartile range for the data set are given as follows:

  • Minimum: 24.
  • Lower quartile: 29.
  • Median: 43.
  • Upper quartile: 50.
  • Maximum: 56.
  • Interquartile range: 50 - 29 = 21.

<h3>What are the median and the quartiles of a data-set?</h3>

  • The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.
  • The first quartile is the median of the first half of the data-set.
  • The third quartile is the median of the second half of the data-set.
  • The interquartile range is the difference between the third quartile and the first quartile.

In this problem, we have that:

  • The minimum value is the smallest value, of 24.
  • The maximum value is the smallest value, of 56.
  • Since the data-set has odd cardinality, the median is the middle element, that is, the 7th element, as (13 + 1)/2 = 7, hence the median is of 43.
  • The first quartile is the median of the six elements of the first half, that is, the mean of the third and fourth elements, mean of 29 and 29, hence 29.
  • The third quartile is the median of the six elements of the second half, that is, the mean of the third and fourth elements of the second half, mean of 49 and 51, hence 50.
  • The interquartile range is of 50 - 29 = 21.

More can be learned about five number summaries at brainly.com/question/17110151

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A = 6 cm; B = 3 cm; C = 8 cm; D = 10 cm

The Surface area of the prism = 120 cm².

<h3>How to Find the Surface Area of Triangular Prism?</h3>

Surface area = Area of A + B + C + D

The side lengths are:

A = 6 cm

B = 3 cm

C = 8 cm

D = 10 cm

Surface area of the prism = 2(area of triangular face) + area of rectangle 1 + area of rectangle 2 + area of rectangle 3

Area of triangular face = 1/2(b)(h) = 1/2(8)(6) = 24 cm

Area of rectangle 1 = (length)(width) = (10)(3) = 30 cm²

Area of rectangle 2 = (length)(width) = (6)(3) = 18 cm²

Area of rectangle 3 = (length)(width) = (8)(3) = 24 cm²

Surface area of the prism = 2(24) + 30 + 18 + 24 = 120 cm².

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