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san4es73 [151]
3 years ago
13

What is the response variable in this scenario? A. B. C. D.

Mathematics
2 answers:
Nitella [24]3 years ago
7 0
The answer is a sorry if it’s wrong
Sever21 [200]3 years ago
3 0

Answer:

It's D

Step-by-step explanation:

Got it right edge 2021

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Please answer this. thanksssss
Sunny_sXe [5.5K]

Answer:the answer is A my friend ;)

Step-by-step explanation:

4 0
3 years ago
154 and 145 to the nearest hundred
Yuki888 [10]

Answer: 154: 200 145: 100

Step-by-step explanation:

If the number next to the place value is 5 or more you round it up. Like I did 154. But if the number is less that 5 you put everything to zero EXCEPT for the place value you are rounding. Good Luck

5 0
3 years ago
Read 2 more answers
A fruit company sells two types of boxes large and small. A delivery of two large boxes and three small boxes has a total weight
lara31 [8.8K]

Answer:

the large box weight is 6.50 kg and the small box weight is 13.75 kg

Step-by-step explanation:

The computation of each type of box weight is as follows:

Let us assume the large box be x

And, the small box be y

So,

2x + 3y = 47.......(i)

6x + 5y = 115........(ii)

Multiply by 3 in equation 1

6x + 9y = 141

6x + 5y = 115

Now subtract the last equation from the above one

4y = 26

y = 6.50

For x, it would be

2x+ 3(6.50) = 47

2x + 19.5 = 47

2x = 47 - 19.50

2x = 27.50

x = 13.75

Hence, the large box weight is 6.50 kg and the small box weight is 13.75 kg

4 0
3 years ago
A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row. All the lights start out off. The first
Gnom [1K]

<u>Solution-</u>

A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row.

As all the lights start out off, in the first pass all bulbs will be turned on.

In the second pass all the multiples of 2 will be off and rest will be turned on.

In the third pass all the multiples of 3 will be off, but the common multiple of 2 and 3 will be on along with the rest. i.e all the multiples of 6 will be turned on along with the rest.

In the fourth pass 4th light bulb will be turned on and so does all the multiples of 4.

But, in the sixth pass the 6th light bulb will be turned off as it was on after the third pass.

This pattern can observed that when a number has odd number of factors then only it can stay on till the last pass.

1 = 1

2 = 1, 2

3 = 1, 3

<u>4 = 1, 2, 4</u>

5 = 1, 5

6 = 1, 2, 3, 6

7 = 1, 7

8 = 1, 2, 4, 8

9 = 1, 3, 9

10 = 1, 2, 5, 10

11 = 1, 11

12 = 1, 2, 3, 4, 6, 12

13 = 1, 13

14 = 1, 2, 7, 14

15 = 1, 3, 5, 15

16 = 1, 2, 4, 8, 16

so on.....

The numbers who have odd number of factors are the perfect squares.

So calculating the number of perfect squares upto 1800 will give the number of light bulbs that will stay on.

As,  \sqrt{1800} =42.42  , so 42 perfect squared numbers are there which are less than 1800.

∴ 42 light bulbs will end up in the on position. And there position is given in the attached table.

7 0
3 years ago
Help me plz !!!!!!!!!!​
kiruha [24]

Answer:

The answer is b. 72.3cm

Step-by-step explanation:

11.5cm is the radius

To find the circumference multiply the radius by 2 then multiply it again by 3.14

3.14 is the beginning of pi.

e.g - 2 × 3.14 x 11.5

5 0
3 years ago
Read 2 more answers
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