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Elenna [48]
3 years ago
7

Jason tossed a fair coin 3 times. What is the probability of getting a head and two tails in any order?

Mathematics
2 answers:
mafiozo [28]3 years ago
4 0
The correct answer is D 
jek_recluse [69]3 years ago
4 0
Hello there.

Question: <span>Jason tossed a fair coin 3 times. What is the probability of getting a head and two tails in any order? 

A. 3/8
B. 4/8
C. 5/8
D. 6/8

Answer: It would be A. 3/8

Hope This helps You!
Good Luck Studying ^-^</span>
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5 0
2 years ago
6. a. Sixty students in a class took an examination in Physics and Mathematics. If 17 of them passed Physics only, 25 passed in
Ivahew [28]

Let C be the set of all students in the <u>c</u>lassroom.

Let P and M be the sets of students that pass <u>p</u>hysics and <u>m</u>ath, respectively.

We're given

n(C) = 60

n(P \cap M') = 17

n(P \cap M) = 25

n((P \cup M)') = n(P' \cap M') = 9

i. We can split up P into subsets of students that pass both physics and math (P\cap M) and those that pass only physics (P\cap M'). These sets are disjoint, so

n(P) = n(P\cap M) + n(P\cap M') = 25 + 17 = \boxed{42}

ii. 9 students fails both subjects, so we find

n(C) = n(P\cup M) + n(P\cup M)' \implies n(P\cup M) = 60 - 9 = 51

By the inclusion/exclusion principle,

n(P\cup M) = n(P) + n(M) - n(P\cap M)

Using the result from part (i), we have

n(M) = 51 - 42 + 25 = 34

and so the probability of selecting a student from this set is

\mathrm{Pr}(M) = \dfrac{34}{60} = \boxed{\dfrac{17}{30}}

7 0
2 years ago
A shop is having a sale.
labwork [276]

Answer:

77

Step-by-step explanation:

10%=11

because you divide 110 by 10.

Then i would multiply that by 3, giving you 33

110-33=77

8 0
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