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natima [27]
3 years ago
14

Estimate 7956 - 5853 by first rounding each number to the nearest hundred.

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

2100

Step-by-step explanation:

Given the following question:

7956-5853

To estimate by the nearest hundred, we have to find the hundreds place value and then look at the number next to it, to find out if that number is greater than or equal to five.

7956-5853

7956

5=5

8000

5853

5=5

5900

8000-5900=2100

=2100

Hope this helps.

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Which of the following are solutions to the equation below?
Luden [163]

Answer:

c) -4/5

d) 4/5

Step-by-step explanation:

The given equation is 25x^2 -16 = 0

Adding 16 on both sides, we get

25x^2 = 16

Dividing both sides by 25, we get

x^2 = 16/25

Taking square root on both sides, we get

x = √(16/25)

x = ±4/5

Therefore, x = 4/5 and x = -4/5

Answers: C) and D)

Hope this will helpful to understand the concept.

Thank you.

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The value of the 4th term is 80. the sequence is being doubled each step what is the 7th term?
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an = ar^(n - 1)
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Step-by-step explanation:

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Read 2 more answers
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}.

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3 years ago
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