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iragen [17]
3 years ago
13

A bag with 8 marbles is shown below. (2 marbles are yellow, 3 are red, and 3 are blue.) A marble is chosen from the bag at rando

m. What is the
probability that it is yellow or red?
Write your answer as a fraction or a whole number.

Mathematics
2 answers:
Andreas93 [3]3 years ago
7 0

Answer:

\frac{5}{8}

Step-by-step explanation:

Hi there!

We're given that there are 8 marbles in total. 2 marbles are yellow and 3 are red, making the number of red AND yellow marbles 5 (2+3=5)

Because 5 out of the 8 marbles are yellow or red, the probability of choosing one of them is \frac{5}{8}.

I hope this helps!

mrs_skeptik [129]3 years ago
7 0

Answer:

P(Red or Yellow) = 5/8

Step-by-step explanation:

2 (yellow) + 3 (red) = 5

Hence, the probability is 5/8

<em>Feel free to mark it as brainliest :D</em>

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Is y=x squared +2 a linear function
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Select the two expressions that are equivalent to 4 + w + 12w.
Black_prince [1.1K]

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A. 4+ 13w and C. 2(2+6w) + w

Step-by-step explanation:

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6 0
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zhuklara [117]

Answer:

3x²-29x+4

Step-by-step explanation:

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4 0
2 years ago
Find the greatest number that divides451, 328 and 673 leaving 1, 4 and 1,respectively as remainders.
Vesna [10]

You're looking for the largest number <em>x</em> such that

<em>x</em> ≡ 1 (mod 451)

<em>x</em> ≡ 4 (mod 328)

<em>x</em> ≡ 1 (mod 673)

Recall that

<em>x</em> ≡ <em>a</em> (mod <em>m</em>)

<em>x</em> ≡ <em>b</em> (mod <em>n</em>)

is solvable only when <em>a</em> ≡ <em>b</em> (mod gcd(<em>m</em>, <em>n</em>)). But this is not the case here; with <em>m</em> = 451 and <em>n</em> = 328, we have gcd(<em>m</em>, <em>n</em>) = 41, and clearly

1 ≡ 4 (mod 41)

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So there is no such number.

4 0
3 years ago
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