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trasher [3.6K]
2 years ago
11

Jackson is playing a game of chance where he must randomly draw a marble out of a jar. He calculates the probability of drawing

a red marble to be 3/15 . What is the probability of not drawing a red marble?
Mathematics
1 answer:
madam [21]2 years ago
5 0

Answer:

12/15

Step-by-step explanation:

If the probability of drawing a read marble is 3/15

Then that would just make the probability of NOT drawing a red marble 12/5

You might be interested in
Evaluate the expression 3÷1 1/5+1/2
Soloha48 [4]
 61
----
110

The answer is 61 over 110, hope this helped you.

8 0
3 years ago
Calculate the total surface area of the pyramid. Show your work!
Vinil7 [7]

Answer: 22.44

Step-by-step explanation:

Base × Height /2 → (3 × 2.24)/2

3.36 · 4 = 13.44

13.44 + 3²

22.44

7 0
3 years ago
ZA and B are complementary angles. If m_A = (x – 15)° and m ZB = (x + 3)º, then find the measure of ZB.​
BlackZzzverrR [31]

Answer:

ZB = 54

Step-by-step explanation:

Complementary angles add to 90 degrees

ZA + ZB = 90

x-15 + x+3 = 90

Combine like terms

2x-12 =90

Add 12 to each side

2x-12+12 = 90+12

2x = 102

Divide by 2

2x/2 = 102/2

x = 51

We want to find the measure of ZB

ZB = x+3

ZB = 51+3

ZB = 54

8 0
2 years ago
At what value of x does the graph of the following function f(x) have a vertical asymptote? f(x)=1/x+3
ratelena [41]

Answer:

d

Step-by-step explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined.

Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non zero for this value then it is a vertical asymptote.

x + 3 = 0 ⇒ x = - 3 ← equation of vertical asymptote

5 0
3 years ago
Read 2 more answers
Anyone know I’m having a hard time!
Rudiy27

Answer:

x = \sqrt{30}

Step-by-step explanation:

Applying altitude- on - hypotenuse theorem.

(leg of large Δ )² = ( part of hypotenuse below it ) × ( whole hypotenuse )

x² = 3 × (3 + 7) = 3 × 10 = 30 ( take square root of both sides )

x = \sqrt{30}

5 0
3 years ago
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