Item 22

Mathematics

1 answer:

Ivan3 years ago

5 0

Answer:

five and three fourths

Step-by-step explanation:

eight ounces is 0.5 of a pound with is half of a pound

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Someone please.. help me in this please I don’t want to fail it..please

igomit [66]

Answer:

So it wants you to first find the mean (average) of the data set, then find the distance everything is from the mean (average).

Step-by-step explanation:

3+5+9+14+16+18= 65

65 divided by the amount of numbers you put in (6) = 10.83 rounded to the nearest hundreth

So now we know the mean (average) now we need to see how far away each number is from the mean (average).10.83-3 is 7.83. 10.83-5 is 5.83. 10.83-9 is 1.83. 10.83-14 is -14.17, but -14.17 would be a normal 14.17 because it wants the absolute value. 10.3-16 is -5.17 (of course keep it as a positive), and finally 10.83-18 is -7.17 as a positive. Now for the mean absolute value of the data they likely mean from 0, so it would be 10.83.

Hope it helped and wasn't too confusing :)

7 0

3 years ago

What is the slop of a line that passes through the points (-3,2) and (-6,5)

denpristay [2]

Answer:
Slope is : -1

8 0

2 years ago

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What is the surface of a box of cereal ( 12in tall and 8in wide

Ipatiy [6.2K]

Answer:

192 inches or 16 feet hope this helps

8 0

3 years ago

Solve the expression (-24) + (+6)

ASHA 777 [7]

Answer:

-18

Step-by-step explanation:

The answer is -18 because a negative plus a positive is always a negative.

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3 years ago

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In the pendulum formula , we use g = 9.8 m/s2 for the acceleration due to gravity on Earth. But what about on the Moon? If an as

Alex Ar [27]

Answer:

The acceleration due to gravity on the moon is 1.642 m/s².

Step-by-step explanation:

Given : In the pendulum formula , we use g = 9.8 m/s2 for the acceleration due to gravity on Earth. If an astronaut on the surface of the moon swings a 1-meter long pendulum, and it has a period of 4.9 seconds.

To find : What is the acceleration due to gravity, g, on the moon?

Solution :

The pendulum formula is $T=2\pi \sqrt{\frac{L}{g}}$