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shepuryov [24]
3 years ago
10

Container A and B hold 11875 L of water altogether. Container B holds 2391 L more than container A holds. How much water does co

ntainer A hold?
Mathematics
1 answer:
vampirchik [111]3 years ago
6 0
Container A holds 9484 L of water
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According to these three facts, which statements are true? - Circle D has center (2, 3) and radius 7. - Circle F is a translatio
KengaRu [80]
B AND C  because its talking about circle

  
4 0
4 years ago
Write at least three statements showing what you know about the fraction 2/3.
Oksi-84 [34.3K]

-The decimal is .66 repeated

- This is a proper fraction, meaning it doesn't have a whole number next to it.

- This isn't a whole number

6 0
3 years ago
PLAZ HURI AND ANSOR WORT 50 POITNZS AND MAOR IF U GET BRIANLEUSTSTS!!!! (Last one for today lul)
klemol [59]
My best educated guess would be D.  
4 0
4 years ago
Read 2 more answers
Can a triangle have side measurements of 1, 4 and 5? Why or why not
Vladimir79 [104]

Answer:

A triangle cannot have side measurements of 1, 4 and 5.

As the sum of two sides i.e. '1+4' of a triangle is not greater than the measure of the third side i.e. '5'.

Step-by-step explanation:

The Triangle Inequality Theorem states that the sum of any two sides of a triangle must be greater than the measure of the third side.

For example, a triangle ΔABC must follow the three conditions which are as follows:

  • A + B > C
  • B + C > A
  • A + C > B

For example, a triangle ΔABC with side lengths:

A = 1

B = 4

C = 5

  • A + B > C    →   1 + 4 > 5      (FALSE)
  • B + C > A    →   4 + 5 > 1      (TRUE)
  • A + C > B    →    1 + 5 > 4     (TRUE)

As the condition i.e. A + B > C    →   1 + 4 > 5  is not satisfied, as 1 + 4 > 5 is False.

Therefore, a triangle cannot have side measurements of 1, 4 and 5.

7 0
3 years ago
A bookstore classifies its books by reader group, type of book, and cost. What is the probability that a book selected at random
Murrr4er [49]

Answer:

Pr(Cb) & Pr(cost $15) = 15(Ncb)/(T1×T3)

Step-by-step explanation:

Classification of the book in to 3

First let's assign terms to each probability

Let:

T1= total number of type of Reader's group

T2 = total number of Type of book

T3 = total number of type of Cost

Probability of child's book and cost $15= Pr(Cb) & Pr(cost $15)

Then find probability of child's book and cost separately:

Pr of child's book = Pr(Cb)

Pr(Cb) = (number of child's book)/ total number of type of group

Ncb = number of child's book

Pr(Cb) = Ncb/T1

Pr (cost $15) = $15 book/total number of cost

Pr (cost $15) = 15/T3

probability of child's book and cost $15 = Pr(Cb) × Pr (cost $15)

= Ncb/T1 × 15/T3

Pr(Cb) & Pr(cost $15) = 15(Ncb)/(T1×T3)

3 0
3 years ago
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