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Sliva [168]
3 years ago
8

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 47.0 and 52.

0 minutes. Find the probability that a given class period runs between 50.25 and 51.0 minutes.
Mathematics
1 answer:
Pavlova-9 [17]3 years ago
7 0

Answer:

0.15 or 15%

Step-by-step explanation:

Since lengths are uniformly distributed, the probability that a class period runs between a two exact times is:

P(x_1\leq x \leq x_2)=\frac{x_2-x_1}{b-a}

In this case, a = 47.0 and b = 52.0 minutes.

The probability that a given class period runs between 50.25 and 51.0 minutes is:

P(50.25\leq x \leq 51.0)=\frac{51.0-50.25}{52-47} \\P(50.25\leq x \leq 51.0)=0.15=15\%

The probability is 0.15 or 15%.

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Thirteen people on a softball team show up for a game. a)How many ways are there to choose 10 players to take the field? b)How m
marishachu [46]

Answer:

a) 286

b) 1,037,836,800

c) 285

Step-by-step explanation:

a) Since order is not important, the total possible number of ways to choose 10 players out of 13 is the following combination:

C(13,10)=\frac{13!}{(13-10)!10! } \\C(13,10)=\frac{13*12*11*10!}{(3*2*1)10! }\\C(13,10) = 286

b) The total number of possibilities to assign positions by the selecting 10 players is the permutation of 13 players for 10 positions:

P(13,10) = \frac{13!}{(13-10)!}\\P(13,10) = 13*12*11*10*9*8*7*6*5*4\\P(13,10) =1,037,836,800

c) The number of ways to pick 10 players including at least one woman is equal to the total number of ways to pick 10 players (found in item a) minus the the number of ways to pick 10 players without picking a single woman.

Since there 10 male players for 10 positions, there is only one possible way to pick a team without women, therefore:

P=286-1 =285

6 0
3 years ago
Is this correct??<br>I don't know
Alexandra [31]
In my opinion, ur right
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A meal coats $16.64. What's the total cost with a 15% tip?​
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Step-by-step explanation:

if 100% ----> $16.64

then 115% ----> $19.136

hope this helps :)

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