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4 years ago

Use the number line to determine the sum −6+4.

Mathematics

2 answers:

Tresset [83]4 years ago

7 0

Answer: -2

Step-by-step explanation: When you're adding and subtracting positives and negatives, it helps to use a number line to visualize what's going on. On a number line, 0 is in the middle and all number to the right of 0 are positive and all numbers to the left of 0 are negative.

When we are simplifying a problem like -6 + 4, we start at 0 and -6 moves us 6 units to the left along our number line. From there, positive 4 then moves us 4 units back to the right and we end up at -2.

This means that -6 + 4 = -2.

Image provided.

N76 [4]4 years ago

3 0

Answer:

-2

Just put the point to -2.

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6. What is the value of x? A. 10 B. 8 C. 4 D. 6

AlladinOne [14]

Answer:

(4)2=x-2

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3 years ago

in a particular class of 26 students , 11 are men . what fraction of the students in the class are women

vitfil [10]

Step-by-step explanation:

woman = total students - men

= 26 - 11

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4 years ago

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Find the probability of randomly rolling 2 standard dice and having a sum that is EVEN or GREATER THAN 9.

Ede4ka [16]

Answer:

Hence, the probability of randomly rolling 2 standard dice and having a sum that is EVEN or GREATER THAN 9 is 2/3

Step-by-step explanation:

Let A be the event that the sum of two dice is even

Then P(A) will be the probability of having even sum

As the sample space is already given in the image

n(S) = 36

A = {(1,1) , (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)}

n(A) = 18

So,

P(A) = n(A)/n(S)

= 18/36

= 1/2

Let B be the event that the sum is greater than 9

B = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}

n(B) = 6

P(B) = n(B)/n(S)

= 6/36

= 1/6

Now we have to find P(A∪B) i.e. P(A or B)

P(A∪B) = P(A) + P(B)

= 1/2 + 1/6

= (3+1)/6

=4/6

=2/3

Hence, the probability of randomly rolling 2 standard dice and having a sum that is EVEN or GREATER THAN 9 is 2/3 ..

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4 years ago

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olga55 [171]

Answer:

${ \tt{2b {}^{3} + 5}} \\ = { \tt{2 {(3)}^{3} + 5 }} \\ = { \tt{59}}$

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3 years ago

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A fair die is tossed once, what is the probability of obtaining neither 5 nor 2?

Leya [2.2K]

Answer:

4/6 or 66.666...%

Step-by-step explanation:

If you want to find the probability of obtaining neither a 5 nor a 2 you find how many times they occur and add them together in this case 5 occurs once and 2 also occurs once out of 6 numbers so 1/6 + 1/6 equals 2/6, you now know that 4/6 of them won't be a 5 nor a 2 and because it is a fair die the likelihood of it falling on a number is the same for all sides so the answer is 4/6 or 66.67%.

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3 years ago