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3 years ago

7

Tom was selling his old games. He started out with 57 but sold 37 of them. He put the rest in boxes with 5 games in each. How ma

ny boxes did Tom have to use?

1 answer:

eduard3 years ago

8 0

Answer:

Tom started out with 57 games but sold 37 of them.

=> Now, Tom has: 57 - 37 =20 games

He put the rest in boxes with 5 games in each.

=> The number of boxes: 20/5 = 4 box

Hope this helps!

:)

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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

$19900 - 4900 - 2842 + 784 = 12942$ .

8 0

3 years ago

Five members of the soccer team and five members of the track team ran the 100-meter dash. Their times are listed in the table b

Sonbull [250]

Soccer : (12.3 + 13.2 + 12.5 + 11.3 + 14.4) / 5 = 63.7 / 5 = 12.74
track : (12.3 + 11.2 + 11.7 + 12.2 + 13.7) / 5 = 61.1 / 5 = 12.22

difference is : 12.74 - 12.22 = 0.52 <=

3 0

3 years ago

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Can u pls help me with this thanks

aleksandrvk [35]

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7 0

3 years ago

Which is larger?<br> 7% or 0.7 and explain how you know.

disa [49]

They are equal because you can measure the percentage as a decimal with a maximum of 1.

7 0

3 years ago

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Solve using elimination. –8x + 6y = –16 –8x + 9y = 8

stira [4]

I think this is right

5 0

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