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natali 33 [55]
3 years ago
12

Priya asked each of five friends to attempt to throw a ball in a trash can until they succeeded. She recorded the number of unsu

ccessful attempts made by each friend as: 1, 8, 6, 2, 4. PRIYA MADE A MISTAKE: The 8 in the data set should have been 18. How would changing the 8 to 18 affect the MEDIAN of the data set? *
Mathematics
1 answer:
TiliK225 [7]3 years ago
3 0

Answer:

1,2,4,6,8

And the median can be calculated from the value in the position 3th and we got 4

Median =4

And now using the correct dataset is: 1, 18, 6, 2, 4. And if we order the dataset we got:

1,2,4,6,18

And the median for this case again would be the value in the 3th position and we got:

Median =4

So then we can conclude that the median is unaffected by the change of 18 instead of 8

Step-by-step explanation:

For this case we have the original dataset given:

1, 8, 6, 2, 4

And the median for this dataset can be calculated with the following dataset ordered:

1,2,4,6,8

And the median can be calculated from the value in the position 3th and we got 4

Median =4

And now using the correct dataset is: 1, 18, 6, 2, 4. And if we order the dataset we got:

1,2,4,6,18

And the median for this case again would be the value in the 3th position and we got:

Median =4

So then we can conclude that the median is unaffected by the change of 18 instead of 8

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​Cameron, Arthur, and Jamie are playing soccer. Their locations are recorded by a motion tracking system. The grid shows distanc
Luden [163]

Answer:

1. Arthur is about 54 meters away from Cameron.

2. Jamie is about 56 meters away from Cameron.

3. Arthur is closer to Cameron.

Step-by-step explanation:

We have been given the locations of Cameron (70,10), Arthur (20,30) and Jamie (45,60) on the the grid in meters.

1. To find the distances of Arthur and Jamie from Cameron we will use distance formula.

\text{Distance}=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

\text{Distance between Cameron and Arthur}=\sqrt{(70-20)^{2}+(10-30)^{2}}

\text{Distance between Cameron and Arthur}=\sqrt{(50)^{2}+(-20)^{2}}

\text{Distance between Cameron and Arthur}=\sqrt{2500+400}

\text{Distance between Cameron and Arthur}=\sqrt{2900}

\text{Distance between Cameron and Arthur}=53.8516480713450403\approx 54

Therefore, the distance Arthur is about 54 meters away from​ Cameron.

2. Let us find the distance between Cameron and Jamie.

\text{Distance between Cameron and Jamie}=\sqrt{(70-45)^{2}+(10-60)^{2}}

\text{Distance between Cameron and Jamie}=\sqrt{(25)^{2}+(-50)^{2}}

\text{Distance between Cameron and Jamie}=\sqrt{625+2500}

\text{Distance between Cameron and Jamie}=\sqrt{3125}

\text{Distance between Cameron and Jamie}=55.9016994374947424\approx 56

Therefore, Jamie is about 56 meters away from​ Cameron.

3. We can see that 56 is greater than 54, therefore, Arthur is closer to Cameron.


4 0
3 years ago
Read 2 more answers
The typical level of a low tide at a beach is the 0 point on a number line
juin [17]

Answer:

The low tide level is 0, integer as 0

Step-by-step explanation:

The tide is at the numberline point 0, making it at integer 0.

Please let me know if I need to change the answer via comments (pls don't report me)

5 0
3 years ago
Mr. Smith gave a cashier a $50 bills for a purchase of $38.70. The cashier gave him a $10 dollar bill, two $1 bills and three di
Aleksandr-060686 [28]

Answer

no he gave him a dollar extra. if his total is 38.70 and gave 50 if you subtract it would be 11.30 not 12.30

Step-by-step explanation:

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3 years ago
A rhombus has sides 7 cm long. If the longer diagonal is 10 cm, what is the area of the rhombus? Round to the nearest hundredth.
timofeeve [1]

Answer:

75 is the correct answer

6 0
3 years ago
What is (+16) - (+2)?​
Nimfa-mama [501]

Answer:

(+16) - (+2) = 14

Step-by-step explanation:

Hope this helped you!

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