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3 years ago

A background feature called _____ allows you to move the background from the slide border in varying distances by a percentage.

1 answer:

5 0

The correct answer is an offset. A background feature called an offset allows you to move the background from the slide border in varying distances by a percentage. The offset is to balance the influences against the opposing influence to get the accurate target.

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3 years ago

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1.) Kilogram- kg

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4 years ago

a 160.-kilogram space vehicle is traveling along astraight line at a constant speed of 800. metersper second. the magnitude of t

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3 years ago

Suppose the internal energy of an ideal gas rises by 2.89 103 J at a constant pressure of 1.00 105 Pa, while the system gains 6.

vampirchik [111]

Answer:

$\Delta V=35.5\ L$

Explanation:

According to the first law of thermodynamics:-

$\Delta U = q - w$

Where,

U is the internal energy = $2.89\times 10^3\ J$

q is the heat = $6.44\times 10^3\ J$ (heat is gained)

w is the work done = ?

$\Delta U = q - w$

$w=q-\Delta U = 6.44\times 10^3\ J-2.89\times 10^3\ J= 3.55\times 10^3\ J$

The expression for the calculation of work done is shown below as:

$w=P\times \Delta V$

Where, P is the pressure, $1.00\times 10^5\ Pa$

$\Delta V$ is the change in volume

Also, 1 J = 1000 PaL

So,

$w=3.55\times 10^6\ PaL$

From the question,

$3.55\times 10^6=1.00\times 10^5\times \Delta V$

$\Delta V=35.5\ L$

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4 years ago

The data represents the daily rainfall (in inches) for one month. Construct a frequency distribution beginning with a lower cla

gladu [14]

The question is incomplete! Complete question along with step by step explanation is provided below!

Question:

The data represents the daily rainfall (in inches) for one month. Construct a frequency distribution beginning with a lower class limit of 0.00 and use a class width of 0.20. Does the frequency distribution appear to be roughly a normal distribution?

0.48 0 0.09 0.17 0 0 0.21 0 0.59 0.13 0 0 0.01 0 0 0.23 0 0 0.01 0 0 1.24 0.24 0 0.05 0.65 0 0 0.28 0

Answer:

No, the given frequency distribution does not appear to be a normal distribution.

Step by Step Explanation:

Monthly rainfall data is given each value represents 1 day and total data points are 30.

0.48 0 0.09 0.17 0 0 0.21 0 0.59 0.13 0 0 0.01 0 0 0.23 0 0 0.01 0 0 1.24 0.24 0 0.05 0.65 0 0 0.28 0

First we will construct a frequency distribution stating at 0.00 with 0.20 width and ending at 1.40.

Range | Frequency

0.00 - 0.20 | 22

0.20 - 0.40 | 4

0.40 - 0.60 | 2

0.60 - 0.80 | 1

0.80 - 1.00 | 0

1.00 - 1.20 | 0

1.20 - 1.40 | 1

Mean = (0.10*22+0.30*4+0.50*2+0.70*1+1.3*1)/30 = 0.21

where midpoint is calculate using (lower limit + upper limit)/2

In a normal distribution, frequencies start off low then increase gradually then decrease gradually and finally end off low.

But in this distribution, as you can see frequencies doesn't start off low but rather the opposite is true. The distribution is very much skewed at the beginning since mean is 0.21 and also the distribution is not symmetric.

Conclusion:

Hence we can conclude that the given frequency distribution does not appear to be a normal distribution.

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3 years ago