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laiz [17]
3 years ago
8

When Jeremy surveyed thirty 7th-grade students at the community pool, he found that 2/15 had not read at least one book during J

uly. He knows there are about 200 7th-grade students in his community, so he infers that about 40 of them did not read any books in July. Is that a good inference? Explain.
Mathematics
2 answers:
WITCHER [35]3 years ago
4 0

.....Jeremy did not make a good inference. The ratio is not in proportion to 2/15. The correct inference would be about 26 students who did not read any books in July. The sample might be biased since it was limited to those at the pool....

Valentin [98]3 years ago
3 0

Answer:

Jeremy did not make a good inference. The ratio is not in proportion to 2/15. The correct inference would be about 26 students who did not read any books in July. The sample is biased since it was limited to those at the pool

Step-by-step explanation:

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Sorry I need help with these 3! Thank you to whoever takes the time to solve them!!!
mash [69]
1.)
Starts with 40:10, then cuz you have 80 buns, you need to x2, so it would be 20 cubes

2.)
I don’t know, sorry

3.)
4:2 and 60 in total
so 4+2 is 6 and 60/6 is 10
so it would be 40:20 so 20

hope this helps :)
8 0
3 years ago
The perimeter of a rectangular field is 380 yd. the length is 70 yd longer than the width. find the dimensions.
GaryK [48]
P=2L+2W
380=2*70+2W
380=140+2W
2W=240
W=120

The dimensions are 70 yd and 120 yd

Hope this helps!
3 0
3 years ago
Fifty six plus blank equals five hundred and four six
anzhelika [568]

Answer:

490

Step-by-step explanation:

546-56= 490

490+56= 546

4 0
3 years ago
Read 2 more answers
I need help hdbcjsjfnfnfjfjffdbbfdncndnc
lbvjy [14]
\frac{3}{15k+30} : \frac{6k}{5k^2+30k+40} = \frac{3}{15(k+2)} . \frac{5(k+2)(k+4)}{6k} = \frac{k+4}{6k}
8 0
3 years ago
The frequency table shows how many visits to the library students in a school have made in the last month.
vfiekz [6]
<h3>Answer:  2.8</h3>

=======================================================

Explanation:

Multiply each visit count with their corresponding frequency.

Examples:

  • 0*12 = 0 for the first row.
  • 1*366 = 366 for the second row
  • 2*53 = 106 for the third row

and so on...

I recommend making a third column like this

\begin{array}{|c|c|c|} \cline{1-3}\text{Visits} & \text{Frequency} & \text{Visits*Frequency}\\\cline{1-3}0 & 12 & 0\\\cline{1-3}1 & 366 & 366\\\cline{1-3}2 & 53 & 106\\\cline{1-3}3 & 52 & 156\\\cline{1-3}4 & 155 & 620\\\cline{1-3}5 & 243 & 1215\\\cline{1-3}\end{array}

That way you can keep track of all the results in an organized way.

Then add everything in the third column

0+366+106+156+620+1215 = 2463

Divide this sum over the total frequency (12+366+53+52+155+243 = 881) and we'll get the mean

2463/881 = 2.7956867

Rounding to one decimal place gets us to 2.8 as the final answer.

-------------

The much longer way to do this is to imagine 12 copies of "0", 366 copies of "1", 53 copies of "2", and so on. We'll have an extremely large data set of 881 items inside it. As you can see, this second method is definitely not recommended to actually carry out. Rather it's helpful to have this as a thought experiment to see why we revert to multiplication instead.

Eg: Imagine adding 155 copies of "4". A shortcut is to simply say 4*155 = 620

4 0
1 year ago
Read 2 more answers
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