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spayn [35]
3 years ago
14

24+4(3+1) answer plz

Mathematics
2 answers:
Gekata [30.6K]3 years ago
7 0

Answer:

40

Step-by-step explanation:

follow the BEDMAS

24+4(3+1)

=24+4(4)

=24+16

=40

Arlecino [84]3 years ago
5 0

Answer:

40

Step-by-step explanation:

24+4(3+1) Order of operations(Bedmas)

24+4(3+1)

24 + 12 + 4. Left to right

36 + 4

40

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Which three comparisons are true?
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Answer:

The answer is 4/8=2/4 because 4/8 simplifying is 2/4 or 1/2

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What quadrant is point (-9,2) located in?​
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If ε = {whole numbers less than 50 but greater than 20} and X = {perfect squares}, Y = {factors of 12}, Z = {prime numbers}; fin
Zigmanuir [339]

Answer:

A. X∪Y = {25, 36, 49}

B. X∩Y = ∅ i.e empty set

C. X' = {21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48}

D. X′∩Y∩Z = ∅

Step-by-step explanation:

We'll begin by determining the universal set (ε), set X, set Y and set Z.

This can be obtained as follow:

ε = {whole numbers less than 50 but greater than 20}

ε = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49}

X = {perfect squares}

X = {25, 36, 49}

Y = {factors of 12}

Y = ∅ i.e empty

Z = {prime numbers}

Z = {23, 29, 31, 37, 41, 43, 47}

A. Determination of X∪Y

X = {25, 36, 49}

Y = ∅

X∪Y =?

X∪Y => combination of elements in set X and Y without repeating any element in both X and Y.

X∪Y = {25, 36, 49}

B. Determination of X∩Y

X = {25, 36, 49}

Y = ∅

X∩Y =?

X∩Y => elements common to both set X and Y

X∩Y = ∅ i.e empty

C. Determination of X′

ε = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49}

X = {25, 36, 49}

X' =?

X' => elements in the universal set but not found in set X.

X' = {21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48}

D. Determination of X′∩Y∩Z

X' = {21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48}

Y = ∅

Z = {23, 29, 31, 37, 41, 43, 47}

X′∩Y∩Z =?

X′∩Y∩Z => elements common to set X', Y and Z

X′∩Y∩Z = ∅

4 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
ANEK [815]

Answer:

Step-by-step explanation:

The probability of picking a red candy from a full bag is 0.20.

But we also have experimental information represented by those 20 digits, each of which tells us how many red candies are in each package.

In the 2nd package there is 1 red candy.  In the fourth there is 1 red candy (since 2 represents a red candy, just like 1 represents a red candy).

Next, in the 6th and 8th packages there is 1 red candy each.  Finally, in the 19th package there is 1 red candy.    Now add up these results:  the number of red candies is 1 + 1 + 1 + 1 + 1, or 5.  This seems to indicate that there will be 5 red candies in the entire 20 packages.  

Caution:  What I have shared here is MY personal interpretation of what we are being asked.  

5 0
4 years ago
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