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Stolb23 [73]
3 years ago
13

In a marble collection, 1/8 of the marbles are blue. Of the blue marbles, 1/2 have sparkles. What fraction of the marbles in the

collection are blue with sparkles?
A. 1/12
B. 4
C. 1/16
D. 5/8
Mathematics
1 answer:
Delvig [45]3 years ago
8 0

Answer:

1/16

Step-by-step explanation:

if half the amount of blue marbles have sparkles you need to divide the amount of blue marbles in half

so 1/2 of 1/8 is 1/16

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What is the probability of choosing a clubs or a numbered card from a deck of cards? Type your answer as a fraction.
Vlada [557]

Answer:

13/52

Step-by-step explanation:

8 0
3 years ago
The four-member math team at Pecanridge Middle School is chosen from the math club, which has three girls and five boys. How man
cestrela7 [59]

Answer:

Total\ Selection = 30\ ways

Step-by-step explanation:

Given

Girls = 3

Boys = 5

Required

How many ways can 2 boys and girls be chosen?

The keyword in the question is chosen;

This implies combination and will be calculated as thus;

Selection =\  ^nC_r = \frac{n!}{(n-r)!r!}

For Boys;

n = 5 and r = 2

Selection =\  ^5C_2

Selection = \frac{5!}{(5-2)!2!}

Selection = \frac{5!}{3!2!}

Selection = \frac{5 * 4 * 3!}{3!*2 * 1}

Selection = \frac{20}{2}

Selection = 10

For Girls;

n = 3 and r = 2

Selection =\  ^3C_2

Selection = \frac{3!}{(3-2)!2!}

Selection = \frac{3!}{1!2!}

Selection = \frac{3 * 2!}{1 *2!}

Selection = \frac{3}{1}

Selection = 3

Total Selection is calculated as thus;

Total\ Selection = Boys\ Selection * Girls\ Selection

Total\ Selection = 10 * 3

Total\ Selection = 30\ ways

5 0
4 years ago
Venla is 5 years older than her cousin Kora. Write an equation for the age of Venla, v, when Kora is k years old.
Talja [164]
K + 5 =V
Kiena + 5 years = Venus’s age
5 0
4 years ago
Read 2 more answers
Identify the terms and the like terms in this expression.
adoni [48]

Answer:

b

Step-by-step explanation:

8 0
3 years ago
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Prove:
Levart [38]
Given the statement, "If n^2 is odd, then n is odd," its contrapositive claims that, "If n is not odd, then n^2 is not odd."

So assume n is not odd, i.e. n is even. This means there is an integer k for which n=2k. Squaring this gives n^2=(2k)^2=4k^2.

Well, we can write 4k^2=2(2k^2), and 2k^2 is just another integer, which means 4k^2=(2k)^2=n^2 must be even.
3 0
4 years ago
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