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3 years ago

What is value of the following decimal as a fraction

1 answer:

7 0

Step 1: Write down the decimal divided by 1, like this: decimal 1.
Step 2: Multiply both top and bottom by 10 for every number after the decimal point. ...
Step 3: Simplify (or reduce) the fraction.

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Maksim231197 [3]

Answer:

D. x=1

Step-by-step explanation:

Image can be rewritten as 3x + 6 = 9.

Solve for x.

3x +6 = 9

Isolate x by subtracting 6 from both sides.

3x = 3

Get rid of the 3 by dividing by 3 on both sides.

x = 1

Hope this helped.

8 0

2 years ago

Does anyone know what the answer is

Natali5045456 [20]

Answer:

cheating in entrance

Step-by-step explanation:

circumference: 14π

area: 49π

8 0

3 years ago

Read 2 more answers

An elevator is rated to safely hold ten 250 lb people. The elevator is loaded as follows: three 600 lbs cartons of books, Jim (1

iris [78.8K]

To do this, we will need to find how much the elevator can hold in all. Multiply 10 by 250 to get 2500. Since it can also hold 10% more, multiply 2500 and 0.10 to find out how much 10% is. You get 250. Add 2500 and 250 to get how much the elevator can safely hold. It would be 2750.

To find out how much more weight can be added, first find out how much weight is on the elevator collectively. (books) 600+ (Jim) 180+ (Gavin) 330 + (Deb) 250 + (Susan) 120 = 1480. Subtract this from the amount the elevator can hold, and you get how much more weight can be added. 2750-1480= 1270.

The elevator can hold 1270 more ibs.

6 0

3 years ago

Read 2 more answers

Lines g,h, and i are all parallel and m------->1 =55*

Hunter-Best [27]

It would be C:55,
hope this helps u out

7 0

3 years ago

The heights of young women follow a Normal distribution with a mean u = 64.5 inches and standard deviation o = 2.5 inches. The p

mash [69]

Answer:

The heights of young women follow a Normal distribution with a mean u = 64.5 inches and standard deviation o = 2.5 inches. The probability that the mean height of an SRS of 10 young women exceeds 66.5 inches is ＿

normalcdf(66.5, 99999999, 64.5, 2.5)

1 - normalcdf(66.5, 99999999, 64.5, 0.7906)

normalcdf(-99999999, 66.5, 64.5, 2.5)

normalcdf(66.5, 99999999, 64.5, 0.7906)

normalcdf(66.5, 99999999, 0, 1)

1 - normalcdf(66.5, 99999999, 64.5, 2.5)

1 - normalcdf(66.5, 99999999, 0, 1)

normalcdf(--99999999, 66.5, 64.5, 0.7906)

plz need help number 1 2 3 4 5 are pic

8 0

2 years ago