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3 years ago

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Mr. Jackson has 30 pennies and 42 dimes that he will be putting into containers for groups of students to use for a math activit

y. He wants to put an equal number of pennies and an equal number of dimes into each container. He will use all the coins.

2 answers:

Ludmilka [50]3 years ago

6 0

if you have 6 groups then there will be 5 pennies and 7 dimes in each group.

Softa [21]3 years ago

5 0

Answer:

<h2>There could be 6 containers with 5 pennies and 7 dime each.</h2>

Step-by-step explanation:

Mr. Jackson whants to put equal numbers of pennies and an equal number of dimes into each container. To know the number of containers could be there, we need to find the greatest common factor between 30 and 42.

Divisors of 30: 1, 2, 3, 5, 6, 10, 15, 30.

Divisors of 42: 1, 2, 3, 6, 7, 14, 21, 42.

Now, observe that the greatest common factor between 30 and 42 is 6, that means there could be a maxium of 6 groups, the number of pennies per group is 30/6=5, and the number of dimes per group is 42/6=7.

Therefore, there could be 6 containers with 5 pennies and 7 dime each.

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Answer:

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Step-by-step explanation:

let the larger number = x

smaller number = y

x - y = 3 1/3 equation 1

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A box contains 3 green marbles, 5 blue marbles, and 7 red marbles. Three marbles are selected at random from the box, one at a t

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Answer:

The probability is $P(K) = \frac{ 28 }{195}$

Step-by-step explanation:

From the question we are told that

The number of green marbles is $n_g = 3$

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The number of red marbles is $n_r = 7$

Generally the total number of marbles is mathematically represented as

$n_t = n_r + n_g + n_ b$

$n_t = 7 + 3 + 5$

$n_t = 5$

Generally total number of marbles that are not red is

$n_k = n_g + n_ b$

=> $n_k = 3 + 5$

=> $n_k = 8$

The probability of the first ball not being red is mathematically represented as

$P(r') = \frac{n_k}{n_t}$

=> $P(r') = \frac{ 8}{15}$

The probability of the second ball not being red is mathematically represented as

$P(r'') = \frac{n_k - 1}{n_t -1}$

=> $P(r'') = \frac{ 8 -1 }{15-1}$ (the subtraction is because the marbles where selected without replacement )

=> $P(r'') = \frac{ 7 }{14}$

The probability that the first two balls is not red is mathematically represented as

$P(R) = P(r') * P(r'')$

=> $P(R) = \frac{ 8}{15} * \frac{ 7 }{14}$

=> $P(R) = \frac{ 8 }{30}$

The probability of the third ball being red is mathematically represented as

$P(r) = \frac{n_r}{ n_t -2}$ (the subtraction is because the marbles where selected without replacement )

$P(r) = \frac{7}{ 15 -2}$

=> $P(r) = \frac{7}{ 13}$

Generally the probability of the first two marble not being red and the third marble being red is mathematically represented as

$P(K) = P(R) * P(r)$

$P(K) = \frac{ 8 }{30} * \frac{7}{ 13}$

=> $P(K) = \frac{ 28 }{195}$

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