Estimating products 3 times 558

Answers

Answer 1

Answer: 600X3=1,800
round to the nearest hundreds 558 rounds up at 600 because 5 or more you round up, 5 or less you round down

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3x + 4y divided by 2 when x=7 and y=10

Which equation is represented by the table? x y (x, y)
0 –1 (0, –1)
1 4 (1, 4)
2 9 (2, 9)

A.
y = 5x – 1

B.
y = –3x + 3

C.
y = 3x – 4

D.
y = 3x + 3

Answers

i would say C.
y= 3x - 4

Which is the least? 3.95, 3 5/6, 3 4/5

Answers

3 4/5
=3.8

3 5/6
=3.8333333333333...

3.95

Therefore, 3 4/5 is the smallest of the three numbers.

3 4/5 : 3.8

3 5/6 : 3.83333333 and so on

3.95 : 3.95

so the answer is 3 4/5

Please answer the question from the attachment.

Answers

15% means 0.15 .

'of' means 'times'.

The problem is (0.15) x (63.2) .

I think you can take over and do it from there.

Tina is fixing to get rectangular signs he wants to space metal trim around the side and just the rectangle measures 32 inches by 9 inches how much trim will Tina need

Answers

You do 32+9=41 times 2 and you get 82 inches.

Karen measured an Orange and a tangerine. The orange Weighed 3/4 pounds. The tangerine weighed 4/6 pounds.which was heavier? Write an inequality to show how their weights compare

Answers

Solutions

Karen measured an Orange and a tangerine. The orange Weighed 3/4 pounds. The tangerine weighed 4/6 pounds. To solve the problem we first have to make the denominators common. To make the denominators common you have to multiply by the GCF ( greatest common factor).

4/6 and 3/4's common denominator is 24

$(4)/(6) = (16)/(24)$

$(3)/(4) = (18)/(24)$

18/24>16/24, the orange is heavier

18/24>16/24, the orange is heavier

Gary bought a gift for his friend and is having it gift wrapped at the store. The wrapping paper comes in four patterns: floral, spiral, cartoon character, and plain. For ribbon, he can choose between red, blue, or green. If Gary randomly selects a paper pattern and a ribbon color, what is the probability that he chooses spiral-patterned paper and a green ribbon?

Answers

Answer: $\text{P(spiral-patterned paper and a green)}=(1)/(12)$

Step-by-step explanation:

Given: The number of patterns in wrapping paper = 4

The number of colors available for ribbon = 3

Then, the total number of types of ribbons are in store = $4*3=12$

Also, the number of type of ribbon has spiral-patterned paper and a green in color =1

So, if Gary randomly selects a paper pattern and a ribbon color, then the probability that he chooses spiral-patterned paper and a green ribbon is

$\text{P(spiral-patterned paper and a green)}=(1)/(12)$

Answer:

Given: The number of patterns in wrapping paper = 4

The number of colors available for ribbon = 3

Then, the total number of types of ribbons are in store =

Also, the number of type of ribbon has spiral-patterned paper and a green in color =1

So, if Gary randomly selects a paper pattern and a ribbon color, then the probability that he chooses spiral-patterned paper and a green ribbon is

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