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satela [25.4K]
3 years ago
12

What is the mode of the data set? 61, 92, 61, 89, 92, 61 Enter your answer in the box.

Mathematics
1 answer:
alisha [4.7K]3 years ago
3 0
61 because it occurs the most
Hope that helped:)!!!!!!!
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The length of a rectangle is 6 cm less than twice its width. Find the dimensions of the rectangle if its areais 108cm
valentinak56 [21]

L=2w-6

area = L x W

108=(2w-6)w

108 = 2w^2-6w

2w^2-6w-108 = 0

2(w^2-3w-54)=0

2(w-9)(w+6) = 0

w=9

L= 2(9)-6 = 12

12*9 = 108

3 0
3 years ago
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
PLEASE HELP ME !!!!!!!!!
Alex777 [14]

Answer:

There are 2 contractions al and del. Name the parts of a verb Ar, Er, ir

3 0
3 years ago
Working together, it takes two different sized hoses 20 minutes to fill a small swimming pool. If it takes 30 minutes for the la
Anuta_ua [19.1K]

Answer:

60 minutes for the larger hose to fill the swimming pool by itself

Step-by-step explanation:

It is given that,

Working together, it takes two different sized hoses 20 minutes to fill a small swimming pool.

takes 30 minutes for the larger hose to fill the swimming pool by itself

Let x be the efficiency to fill the swimming pool by larger hose

and y be the efficiency to fill the swimming pool by larger hose

<u>To find LCM of 20 and 30</u>

LCM (20, 30) = 60

<u>To find the efficiency </u>

Let x be the efficiency to fill the swimming pool by larger hose

and y be the efficiency to fill the swimming pool by larger hose

x = 60/30 =2

x + y = 60 /20 = 3

Therefore efficiency of y = (x + y) - x =3 - 2 = 1

so, time taken to fill the swimming pool by small hose = 60/1 = 60 minutes

8 0
3 years ago
Simplify the following expression:
nalin [4]
-2/2 I think probably
3 0
3 years ago
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