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schepotkina [342]
2 years ago
5

What is the difference in kilowatt hours of monthly energy production between an average local wind speed of 15 miles per hour a

nd an average local wind speed of 10 miles per hour? I really just need a SIMPLE answer! Not an entire break down...
Mathematics
2 answers:
nirvana33 [79]2 years ago
8 0

Answer:

The difference in kilowatt hours of monthly energy production between an average local wind speed of 15 miles per hour and an average local wind speed of 10 miles per hour would be of a third less kilowatts produced with a 10 miles per hour wind than with a 15 miles per hour wind.

Step-by-step explanation:

This is because 10 corresponds to 2/3 of 15, 1/3 being the remaining value of electrical production that would not be generated with a wind of 10 mph rather than a 15 mph wind.

natali 33 [55]2 years ago
6 0

Answer:

the difference is 5

Step-by-step explanation:

difference means subtraction meaning you subtract 15-10 and your answer would be 5 <em>your welcome</em>

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"A consultant compiled the following data set that shows" the number of visits made to the National Museum of American History f
aalyn [17]

Answer:

Step-by-step explanation:

The best option is for the consultant to remove these data points because they are outliers.  Unusual data points which are located far from rest of the data points are known as outliers.

3 0
2 years ago
Alondra is nearly 6 feet tall. Her normal stride is 24 inches long. She wants to walk 5 miles each day for exercise. The table b
Alenkinab [10]

There are 253,440 inches in 4 miles. If you divide this quantity into the 24 inch strides she takes, you get your answer of 10,560 strides.

Hope this helps! <3

8 0
2 years ago
Your tutor asks you to record the time you spend using IT during the week. You have recorded this time below: Day Time Monday 0.
NikAS [45]

Let us convert all figures into decimals so that we can compare them easily.

Monday    0.3

Tuesday   15% = 0.15

Wednesday   1/6 = 0.1666

Thursday   0.2

Friday   1/8 = 0.125

Clearly, I spent the least amount of time on Friday using IT and the time is 0.125 or 1/8.


7 0
2 years ago
Alice and Bob each choose at random a number between zero and two. We assume a uniform probability law under which the probabili
Mnenie [13.5K]

Answer / Step-by-step explanation:

Before we start to answer this question, it would be nice to understand some basic definition and principles:

Random: This refers to an unpredictable pattern of arrangement.  Without a specific aim or logical procedural pattern.

Uniform Probability distribution: This is a type of probability distribution in which the outcomes of distribution irrespective of the pattern or distance results in equally likely outcomes.

Uniform distribution is one of the widely used distributions although it's one of the simplest ones. The idea behind this distribution is that, unlike the other distributions, the likelihood of the any outcome in the distribution range is the same.

Now that we have a brief knowledge of the key terms within the question, hence, to start answering the question,

(a) Starting with 0 for Alice, Bob can choose any number starting from   1 /  3 . As Alice increases, Bob has to increase his starting point as well. Therefore, after Alice passes  1 /  3 , Bob can start choosing numbers from 0. This creates 2 triangles with sides  1  / 3 . The area of them gives us the probability.

Therefore, : P = 2 ( 1/3 1/3 1/3)

                        = 4/9

(b) We can find the probability of having both the numbers less than  1  / 3 . Then we subtract this probability from 1.

We have: P = 1 - 1/3 . 1/3

                     = 8/9

(c) This is basically impossible as the area of a dot is negligible.

(d) It means that Alice needs to select a number between  1  / 3  and 1.

Therefore: P = 1 - 1 /3

                    = 2 / 3

8 0
2 years ago
If you had carried out the algebra using variables before plugging numbers into your expressions, you would have found that (vf)
SashulF [63]

Answer:

Step-by-step explanation:

Volt is energy per electric charge

1V=1J/C

and the energy units in terms of meters, seconds and kilograms are:

1J= 1kg\frac{m^{2}}{s^{2}}

So the unit of volt is

1V= kg \frac{m^{2}}{Cs^{2}}

So the velocity expression (it can not be read completly) but i guess that it could be as follows

vf_{a} = -2\sqrt{\frac{q_{a}\Delta V}{m_{a}}}

So the unit analysis would be

vf_{a} = \sqrt{\frac{C kg\frac{m^{2}}{Cs^{2}}}{kg}}}

note that the coulomb unit of numerator is cancelled from the denominator. The same occurs for kg unit.

vf_{a} = \sqrt{\frac{m^{2}}{s^{2}}}

Applying square root

vf_{a} = \frac{m}{s}

7 0
2 years ago
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