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3 years ago

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You can earn $12.75 an hour working at shop well. You can earn $7.50an hour babysitting on the Weekend.during the month of May ,

you worked hard to earn money for you’re summer vacation you used a table to keep track of the number of hours you worked each day

1 answer:

patriot [66]3 years ago

8 0

where's the question?

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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

Plz help me help me help me

exis [7]

The answer is this :) hope I helped

7 0

2 years ago

Read 2 more answers

A Gardner wants to add 39 pounds of nutrient a and 16 pounds of nutrient b to her garden. Each bag of brand x provides 3 pounds

ehidna [41]

Answer:

5 bags of brand x and 6 bags of brand y.

Step-by-step explanation:

Nutrient requirement for the garden:

39 pounds of nutrient a and

16 pounds of nutrient b

Component of each bag of brand x:

3 pounds of nutrient a and 2 pounds of nutrient b

Therefore, 5 bags of brand x will contain 15 pounds of nutrient a and 10 pounds of nutrient b

Component of each bag of brand y:

4 pounds of nutrient a and 1 pound of nutrient b

Therefore, 6 bags of brand y will contain 24 pounds of nutrient a and 6 pounds of nutrient b

Altogether, she should buy 5 bags of brand x and 6 bags of brand y to meet the nutrient requirement of the garden

3 0

3 years ago

What is 204,902 is scientific notation? also 124,510 in scientific notation.

Nesterboy [21]

204 902.0

124 510.0

Move the decimal point to the left, until you have a number less than 10.

204 902 = 2.04902 × 10⁵

124 510 = 1.2451 × 10⁵

3 0

2 years ago

Read 2 more answers

Maths Problem. I can't quite explain it, could you help please? Thanks for every good answer. All shown on the picture :)

Nimfa-mama [501]

This is something you would do through trial and error. At least, that's the approach I took. I'm not sure if there is any algorithm to solve. The solution I got is shown in the attached image below. There are probably other solutions possible. The trick is to keep each number separate but not too far away so that the other numbers to be filled in later don't get too crowded to their neighbor.

Side note: any mirror copy of what I posted would work as well since you can flip the page around and it's effectively the same solution.

7 0

3 years ago