Distance Formula: 
Apply the points: 
Solve: 
Since the square root of 218 cannot be simplified, the answer is

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Answer:
30 red marbles
Step-by-step explanation:
If the ratio is 2 red marbles for every 5 blue marbles then you would add them together to find out how many marbles are in one group.
2+5=7
There are 7 marbles in each group.
Now we have to find out how many total groups there are by dividing the total amount of marbles by the amount in each group.
105÷7=15
There are 15 groups of 7 marbles.
Looking back at the ratio, we know that the 15 groups each have 2 red marbles and 5 blue marbles. So if there is 2 red marbles 15 times (2 marbles in 15 groups) we would multiply.
2×15=30
There are 30 red marbles.
I hope that makes sense.
Answer:
7. quadrilateral, 8. square, 9. rectangle, 10. rhombus or parallelogram 11. square, 12. square, 13. rhombus or parallelogram, 14. parallelogram or rhombus
Based on the data recorded by Isabella, it can be concluded the cube is rather fair.
<h3>How many times did Isabella get each number?</h3>
Based on the data, here are the results:
- Getting a 1: 6 times
- Getting a 2: 5 times
- Getting a 3: 7 times
- Getting a 4: 5 times
- Getting a 5: 6 times
- Getting a 6: 7 times
This implies, in total Isabella got the same number between five and seven times. For example, the number 2 was obtained 5 times, but the number 3 was obtained 7 times.
<h3>What can be concluded based on the results?</h3>
Even though Isabella did not get the same number of times each number, the dice is rather fair because by rolling the dice thirty six times you will obtain the same number at least five times.
Moreover, there is not a big difference in the number of times you obtain each number.
Learn more about dice in: brainly.com/question/23637540
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Answer:
0.74 to 6.06
Step-by-step explanation:
The groups are independnet,
SE(xh bar-xa bar)=sqrt [sh^2/nh+sa^2/na]=sqrt [10.1^2/80+10.3^2/80]=1.61
At df=157, the t critical is 1.65
90%c.i=(xh bar-xa bar)+-tcritical SE(xh bar-xa bar)
=(25.2-21.8)+-1.65*1.61
=0.74 to 6.06