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dangina [55]
3 years ago
15

Round 80531 to the nearest thousand.

Mathematics
2 answers:
Iteru [2.4K]3 years ago
6 0
That would be 8,1000
Radda [10]3 years ago
5 0

It is 8,0531.I think this answer is correct, if not I am soorry


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state whether each expression is meaningful. if not, explain why. if so, state whether it is a vector or a scalar.
DanielleElmas [232]

The dot product of two vectors gives scalar quantity and between scalar quantity, we can not apply any product so options (b), (d), and (e) are not meaningful.

<h3>What is a vector?</h3>

An item with both magnitude and direction is referred to be a vector.

A vector can be visualized geometrically as a straight spline, with a pointing in the orientation and a length equal to the value of the vector.

A line with an arrow in its direction can also be used to represent a vector, and the length of the line relates to the vector's amplitude.

The dot product of two vectors gives scalar quantity while the cross product gives a vector quantity.

Now,

In option (B) ;

a Х (b . c) now here b . c is a scalar quantity and we can not cross multiply a scalar with a vector.

In option (D) ;

(a.b) Χ (c.d) where we cannot multiply two scalars by cross product.

In option (E);

a.(b.c) we can not multiply two scalars by the dot product.

Hence "The dot product of two vectors gives scalar quantity and between scalar quantity, we can not apply any product".

For more information about vector,

brainly.com/question/13322477

#SPJ4

The given question has a missing option which is given below,

8 0
2 years ago
How many cells are in the human body ​
NARA [144]

Answer:

There should be 37.2 trillion, however this is not a finite number and there might be more or less depending on the person.

Step-by-step explanation:

3 0
3 years ago
Solve it : -<br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B6%20%5Ctimes%2024%7D%20" id="TexFormula1" title=" \sqrt{6 \ti
irga5000 [103]

Answer:

12

Step-by-step explanation:

sqrt(6*24)

sqrt(144)

sqrt(12*12)

12

3 0
3 years ago
Read 2 more answers
Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below: Team X: 11, 3, 0,
Gre4nikov [31]

Answer:

C. Team Y’s scores have a lower mean value.

Step-by-step explanation:

We are given that Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below:

Team X: 11, 3, 0, 0, 2, 0, 6, 4

Team Y: 4, 2, 0, 3, 2, 1, 6, 4

Firstly, we will calculate the mean, median, range and inter-quartile range for Team X;

Mean of Team X data is given by the following formula;

        Mean, \bar X =  \frac{\sum X}{n}

                       =  \frac{11+ 3+ 0+ 0+ 2+ 0+ 6+ 4}{8}  =  \frac{26}{8}  = 3.25

So, the mean of Team X's scores is 3.25.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team X: 0, 0, 0, 2, 3, 4, 6, 11

  • If n is odd, then the formula for calculating median is given by;

                         Median  =  (\frac{n+1}{2} )^{th} \text{ obs.}

  • If n is even, then the formula for calculating median is given by;

                         Median  =  \frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.}  }{2}

Here, the number of observations is even, i.e. n = 8.

So, Median =  \frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.}  }{2}

                   =  \frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.}  }{2}

                   =  \frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.}  }{2}

                   =  \frac{2+3}{2}  = 2.5

So, the median of Team X's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

               Range = Highest value - Lowest value

                           = 11 - 0 = 11

So, the range of Team X's score is 11.

Now, the inter-quartile range of the data is given by;

        Inter-quartile range = Q_3-Q_1

Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}

     =  (\frac{8+1}{4} )^{th} \text{ obs.}

     =  (2.25 )^{th} \text{ obs.}

Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]

     =  0 + 0.25[0 - 0] = 0

Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}

     =  3(\frac{8+1}{4} )^{th} \text{ obs.}

     =  (6.75 )^{th} \text{ obs.}

Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]

     =  4 + 0.75[6 - 4] = 5.5

So, the inter-quartile range of Team X's score is (5.5 - 0) = 5.5.

<u>Now, we will calculate the mean, median, range and inter-quartile range for Team Y;</u>

Mean of Team Y data is given by the following formula;

        Mean, \bar Y =  \frac{\sum Y}{n}

                       =  \frac{4+ 2+ 0+ 3+ 2+ 1+ 6+ 4}{8}  =  \frac{22}{8}  = 2.75

So, the mean of Team Y's scores is 2.75.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team Y: 0, 1, 2, 2, 3, 4, 4, 6

  • If n is odd, then the formula for calculating median is given by;

                         Median  =  (\frac{n+1}{2} )^{th} \text{ obs.}

  • If n is even, then the formula for calculating median is given by;

                         Median  =  \frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.}  }{2}

Here, the number of observations is even, i.e. n = 8.

So, Median =  \frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.}  }{2}

                   =  \frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.}  }{2}

                   =  \frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.}  }{2}

                   =  \frac{2+3}{2}  = 2.5

So, the median of Team Y's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

               Range = Highest value - Lowest value

                           = 6 - 0 = 6

So, the range of Team Y's score is 6.

Now, the inter-quartile range of the data is given by;

        Inter-quartile range = Q_3-Q_1

Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}

     =  (\frac{8+1}{4} )^{th} \text{ obs.}

     =  (2.25 )^{th} \text{ obs.}

Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]

     =  1 + 0.25[2 - 1] = 1.25

Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}

     =  3(\frac{8+1}{4} )^{th} \text{ obs.}

     =  (6.75 )^{th} \text{ obs.}

Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]

     =  4 + 0.75[4 - 4] = 4

So, the inter-quartile range of Team Y's score is (4 - 1.25) = 2.75.

Hence, the correct statement is:

C. Team Y’s scores have a lower mean value.

4 0
3 years ago
Write the value of the digit in the hundreds place and the value of the digit in the tens place in 440. What is the relationship
faltersainse [42]

Both of the digits in the hundreds and tens place are 4, making it a relationship. The 4 in the hundreds place is worth 400, the 4 in the tens place is worth 40, and the 0 in the ones place is worth 0.

Hope this helped!

Nate

4 0
3 years ago
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