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Scilla [17]
4 years ago
6

A group of 10 fisherman enjoy certain types of fish, 3 of them like salmon, 5 like trout and 2 like bass. A single fisherman is

chosen at random. What is the probability that the fisherman likes salmon or bass?
A. 1/10
B. 47/90
C. 3/50
D. 1/2
Mathematics
1 answer:
monitta4 years ago
4 0
It is D, as 2 plus 3 is 5 which is half of 10.
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Select all that are true.
satela [25.4K]

Answer:

1

2

6

Step-by-step explanation:

1/2 +1/2 =1

length 1/2 ×4=2

wide 1/2 ×3 =1 1/2

height 1/2×3= 1 1/2

4 0
3 years ago
Your bank account hos a bolance of -$17. You deposit $70. What is your new balance? Show work:​
Norma-Jean [14]

Answer:

53

Step-by-step explanation:

70-17= 53

3 0
3 years ago
Read 2 more answers
The half-life of a certain radioactive material is 36 days. An initial amount of the material has a mass of 487kg. Write an expo
Lubov Fominskaja [6]

Answer:

442.302 kg

Step-by-step explanation:

amount = 487 (1/2)^(t/36) where t is the number of days.

if t = 5

amount = 487(.5)^(5/36)

= 442.302 kg

4 0
2 years ago
1. Solve the inequality.
Sauron [17]

Answer:

Your final answer is either

x≥-2   if your initial inequality was

6x+2≤2(5-x)

OR

x≤-2

if your initial inequality was

6x+2≥2(x-2)

Step-by-step explanation:

As shown you have an equality, not an inequality.

-6x+2=2(5-x)          distribute through parenthesis

-6x+2=2(5)+2(-x)

-6x+2=10-2x           add 2x to both sides

2x-6x+2=10-2x+2x

-4x+2=10                subtract 2 from both sides

-4x+2-2=10-2

-4x=8                      divide both sides by -4

-4x/(-4) = 8/(-4)

x = -2

With the ≥ or ≤ sign you would solve the exact same way

except for the point where when dividing both sides by

-4 requires you to reverse the inequality.

Your final answer is either

x≥-2   if your initial inequality was

6x+2≤2(5-x)

OR

x≤-2

if your initial inequality was

6x+2≥2(x-2)

7 0
3 years ago
Can you solve this????? Super hard!
Likurg_2 [28]

Answer:

  \textbf{J. }\dfrac{1}{x^2-x}

Step-by-step explanation:

Factor the denominator and cancel the common factor.

  \dfrac{x+1}{x^3-x}=\dfrac{x+1}{x(x^2-1)}=\dfrac{x+1}{x(x-1)(x+1)}=\dfrac{1}{x(x-1)}\\\\=\boxed{\dfrac{1}{x^2-x}}

6 0
3 years ago
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