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Nonamiya [84]
3 years ago
14

Use benchmarks to estimate 2.81+3.73

Mathematics
1 answer:
love history [14]3 years ago
5 0
The benchmarks are:  0,  0.25,  0.50,  0.75  and 1.
   2. 81   →  2.75
+
   3.73    →  3.75
------------
2.75 + 3.75 =  6.50
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Which term can be added to the list so that the greatest common factor of the 3 terms is 12h3
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The term that can be added to the list so the GCF is 12h3 would be 48h5. 
The reason being is that 48 is first divisible by 12 and does not yield a fraction, and we can remove upon dividing 3 h's from this term as it contains a total of 5 h's.
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3 years ago
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Jasmine has 5 books that she wants to read. She wants to read the nonfiction book first and the mystery book last. In how many d
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Need help with this math problem
zheka24 [161]

Answer:

m<1 = 51

m<2 = 61

m<3 = 29

Step-by-step explanation:

three angles of a triangle = 180

so knowing that add 68 + 61 to get 129

now subtract cause the last angle will = 180-129 = 51

for angle two is 61 cause their congruent

now we have one angle = 90 and another 61

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There are 4,389 dogwood trees in the state park. The park workers are going to plant 342 more trees. How many trees will there b
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The table below shows all outcomes for rolling 2 dice. The bold numbers 1 through 6 show the possible results for each die. All
Leokris [45]

Part I:

1. If you have to circle each outcome in the table that at least 1 die is a 4, then you have to circle row corresponding to the number 4 and column corresponding to the number 4.

2. If you have to cross out each outcome where the sum of the dice is greater than 5, then you have to cross out all not bold numbers 6, 7, 8, 9, 10, 11 and 12 in the table.

Part II:

Here you can see 36 possible outcomes.

1. The probability that at least 1 die is a 4 is:

Pr(A)=\dfrac{11}{36} - (favorable outcomes are 4+1=5, 4+2=6, 4+3=7, 4+4=8, 4+5=9, 4+6=10, 1+4=5, 2+4=6, 3+4=7, 5+4=9, 6+4=10).

2. The probability that the sum of the dice is greater than 5 is:

Pr(B)=\dfrac{26}{36}=\dfrac{13}{18}.

3. The probability that at least 1 die is a 4 and that the sum of the dice is greater than 5 is

Pr(A\cap B)=\dfrac{9}{36}=\dfrac{1}{4}.

4. The probability that at least 1 die is a 4 or that the sum of the dice is greater than 5 is:

Pr(A\cup B)=\dfrac{28}{36}=\dfrac{7}{9}.

4 0
3 years ago
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