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ioda
3 years ago
6

Player 1

Mathematics
1 answer:
sveta [45]3 years ago
7 0

Answer:

Player 1's position is Player 2's position reflected across the y-axis; only the signs of the y-coordinates of Player 1 and Player 2 are different

Step-by-step explanation:

You might be interested in
In ΔABC, point M is the midpoint of side AB and point D is the midpoint of segment MC. Prove that the area of ΔADC= the area of
Anna11 [10]
A line segment from a vertex to the midpoint of the opposite side is a "median". A median divides the area of the triangle in half, as it divides the base in half without changing the altitude.
AAMC is half AABC. AADC is half AAMC, so is 1/4 of AABC. (By the formula for area of a triangle.)
ABMC is half AABC. ABMD is half ABMC, so is 1/4 of AABC. (By the formula for area of a triangle.)
Then, AADC = 1/4 AABC = ABMC, so AADC = ABMC by the transitive property of equality.
8 0
3 years ago
The owner expects you to hand wash each coffee pot before the coffee shop opens. It takes 10 minutes for all of the coffee pots
tester [92]

Answer:

Total number of coffee pot = 16 pots

Step-by-step explanation:

Given:

Time taken for drying all coffee pot = 10 minutes

Total time taken for cleaning and drying coffee pot = 82  minutes

Time taken for clean a coffer pot = 4 ½ minutes = 4.5 minutes

Find:

Total number of coffee pot

Computation:

Total time taken for cleaning = Total time taken for cleaning and drying coffee pot  - Time taken for drying all coffee pot

Total time taken for cleaning = 82 - 10

Total time taken for cleaning = 72 minutes

Total number of coffee pot = Total time taken for cleaning / Time taken for clean a coffer pot

Total number of coffee pot = 72 / 4.5

Total number of coffee pot = 16 pots

8 0
3 years ago
HELP ASPAP!!!!! WILL MARK BRAINIEST standard deviation
Umnica [9.8K]

Answer:

The standard deviation of the given data is 11.34.

Step-by-step explanation:

We have to find the standard deviation for the given data;

Grade on     No. of students          

 test (X)        with that score (f)        X \times f       ( X - \bar X)^{2}        f \times ( X - \bar X)^{2}

     50                  1                       50      1073.218      1073.2176

     57                 2                       114      663.578      1327.1552

     60                 4                       240       518.018     2072.0704

     65                 3                       195       315.418      946.2528

     72                 3                       216        115.778       347.3328

     75                12                       900        60.218       722.6112

     77                10                       770        33.178       331.776

     81                 6                       486        3.098       18.5856

     83                 6                       498        0.058        0.3456

     88                9                       792        27.458      247.1184

     90                12                      1080        52.418      629.0112

     92                12                       1104        85.378     1024.5312

     95                 2                        190        149.818      299.6352

     99                 4                        396        263.738     1054.9504

    100          <u>       5   </u>                 <u>    500    </u>     297.218   <u>   1486.088     </u>

   Total             <u>      91       </u>                <u>    7531     </u>                       <u>    11580.68    </u>

Firstly, the mean of the above data is given by;

            Mean, \bar X  =  \frac{\sum X \times f}{\sum f}

                             =  \frac{7531}{91}  =  82.76

Now, the standard deviation of the data is given by;

          Standard deviation, S.D.  =  \sqrt{\frac{\sum f(X-\bar X)^{2} }{\sum f-1}}

                                                     =  \sqrt{\frac{11580.68}{91-1}}

                                                     =  11.34

Hence, the standard deviation of the given data is 11.34.

5 0
3 years ago
The graph compares the weights in pounds of 100 dogs and cats that are brought in to a veterinarian's office.
GenaCL600 [577]
In case of dogs the value 10 is the minimum value. So all the values lie above 10. In total there were 100 dogs.
So for dogs, we can say number of dogs above the value of 10 pound are 100.

In case of Cats, 10 lies at the position of median. Median is the central value and 50% values lie above the median value. So number of cats with weight above 10 pound is 50.

Thus, we can conclude that there were 50 more dogs than the cats with weight over 10 pounds. So option C gives the correct answer.
8 0
3 years ago
Read 2 more answers
The number lxl is the distance between 0 and x it can never equal a ___ number
uysha [10]

Answer:

negative

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
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