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3 years ago

If you divided 37 objects into sets of 5, how many sets of 5 could you make, and how many are left over?

Mathematics

1 answer:

egoroff_w [7]3 years ago

4 0

Answer:

You could make 7 sets and would have 2 left over.

Step-by-step explanation:

5 times 7 equals 35.

If you subttact 37-35 you would get 2.

Therefore you could have 7 sets and 2 left over.

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Which equation represents the line that passes

viktelen [127]

Answer:

1) y=3x+1

Step-by-step explanation:

passes through the points

(-1.-2) and (3, 10)

7 0

3 years ago

Hisako sells dolls at her doll store. If she sells a doll for $35, and there is 6% sales tax, what is the tax amount a customer

Sloan [31]

Given:

The price of a doll = $35

Sales tax = 6%

To find:

The tax amount that a customer will pay on one doll.

Solution:

The price of doll is $35 and the sales tax is 6%, so the tax amount on a doll is 6% of 35.

$\text{Tax amount}=35\times \dfrac{6}{100}$

$\text{Tax amount}=\dfrac{210}{100}$

$\text{Tax amount}=\dfrac{21}{10}$

$\text{Tax amount}=2.1$

Therefore, the tax amount that a customer will pay on one doll is $2.1.

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3 years ago

The students at Southern Junior High School are divided into four homerooms. Benjamin just moved into the school district and wi

brilliants [131]

Answer: A 4 side dice.

Step-by-step explanation:

Benjamin has 4 possible homerooms with the same probability. This means that every homeroom has a 1/4 = 0.25 (or 25%) probability of getting chosen.

You could emulate this situation with a 4 sided dice (are like little pyramids) where each side of the dice would represent the selection of one of the homerooms.

6 0

3 years ago

Please solve the problem with steps

Debora [2.8K]

Answer:

Infinite series equals 4/5

Step-by-step explanation:

Notice that the series can be written as a combination of two geometric series, that can be found independently:

$\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}$

The first one: $(\frac{1}{2})^{n-1}$ is a geometric sequence of first term ( $a_1$ ) "1" and common ratio (r) " $\frac{1}{2}$ ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: $Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2$

The second one: $\frac{1}{6^{n-1}}$ is a geometric sequence of first term "1", and common ratio (r) " $\frac{1}{6}$ ". Again, since the common ratio is smaller than one, we can find its infinite sum: