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3 years ago

Please don’t have a friend

SAT

1 answer:

Licemer1 [7]3 years ago

8 0

Explanation:

what well I'm ur freind now mate

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Why might it be adaptive for stomata to occur mostly (if not entirely) on the undersides of leaves?

adoni [48]

Answer:

because they will recieve less sunlight therefore water will be losed at a slower rate,water does not evaporate as quickly, which allows for transpiration to occur.

Explanation:

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2 years ago

Is this statement true or false? Smallpox weakened the Aztec and Inca Empires. Question 11 options: True False

Lana71 [14]

Answer:

true

Explanation:

bc i took the test

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2 years ago

A student pushes a 51. 5 kilogram bookshelf across a smooth floor with a net force of 67 n.

topjm [15]

Assuming a student pushes a 51. 5 kilogram bookshelf across a smooth floor with a net force of 67 n the acceleration is 1.30m/s².

<h3>Acceleration</h3>

Using this formula

Acceleration= Net force/Mass of bookshelf

Where:

Mass of bookshelf =51.5 kilogram

Net force =67 Newton

Let plug in the formula

Acceleration=67n/51.5 kilogram

Acceleration=1.30 m/s²

Inconclusion the acceleration is 1.30m/s².

Learn more about acceleration here:brainly.com/question/25793230

7 0

2 years ago

To compare the average amount of time that canadians and americans spend commuting, a researcher collects a sample of canadians

yKpoI14uk [10]

The standard error of the difference of sample means is 0.444

From the complete question, we have the following parameters

<u>Canadians</u>

Sample size = 50
Mean = 4.6
Standard deviation = 2.9

<u>Americans</u>

Sample size = 60
Mean = 5.2
Standard deviation = 1.3

The standard error of a sample is the quotient of the standard deviation and the square root of the sample size.

This is represented as:

$SE = \frac{\sigma}{\sqrt n}$

The standard error of the Canadian sample is:

$SE_1 = \frac{2.9}{\sqrt{50}}$

So, we have:

$SE_1 = 0.41$

The standard error of the American sample is:

$SE_2 = \frac{1.3}{\sqrt{60}}$

So, we have:

$SE_2 = 0.17$

The standard error of the difference of sample means is then calculated as:

$SE= \sqrt{SE_1^2 + SE_2^2}$

This gives

$SE= \sqrt{0.41^2 + 0.17^2}$

$SE= \sqrt{0.197}$

Take square roots

$SE= 0.444$

Hence, the standard error of the difference of sample means is 0.444