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-BARSIC- [3]
3 years ago
13

Cooper is blocking off several rooms in a hotel for guests coming to his wedding. The hotel can reserve small rooms, that can ro

om 2 people, and large rooms, that can room 4 people. How many wedding guests could stay in the hotel if Cooper reserved 12 small rooms and 11 large rooms? How many wedding guests could stay in the hotel if Cooper booked ss small rooms and ll large rooms?
Mathematics
1 answer:
e-lub [12.9K]3 years ago
6 0

Answer:

Step-by-step explanation: s = small room = 2

l = large room = 4

12s + 11l = 28 + 22 = 50

as I said before s = 2 and l = 4

s*s + l*l = 2*2 + 4*4 = 4 + 16 = 20

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When a six sided die is rolled then what is the probability that the number<br> rolled will be four?
vazorg [7]

Answer:1/6

Step-by-step explanation:

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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
The diagram shows a rhombus inside a regular hexagon work out the angle x
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Answer:

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Step-by-step explanation:

The rest of the question is the attached figure.

And it is required to find the angle x.

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So, the measure of each acute angle of the rhombus + the measure of angle x = the measure of one angle of the regular hexagon.

So,

60 + x = 120

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<u>So, the measure of the angle x = 60°</u>

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