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aleksandr82 [10.1K]
4 years ago
12

Find the distance between points P(8, 2) and Q(3, 8) to the nearest tenth.

Mathematics
1 answer:
riadik2000 [5.3K]4 years ago
7 0
<span>First of all to calculate the distance between two points we can use distance formula
   
d=Square Root [(x2-x1)^2 + (y2-y1)^2]
   
Now substitute the given points p(x1,y1) and q(x2,y2)in above distance formula
   
The values are X2=3, X1=8and Y2=8and Y1=2.
   
After Substituting the values
   
d=Square Root[(-5)^2+(6)^2]
   
d=Square Root(25+36]
   
d=Square Root[61]
   
d=7.8
   
7.8 is the distance between points P(8, 2) and Q(3, 8) to the nearest tenth.</span>
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<em>70%</em>

<em></em>

Step-by-step explanation:

Given

Number of Siblings:   ||    0  ||    1    ||      2     ||       3

Number of Students: ||    4  ||    18    ||    10     ||       8

Required

Determine the probability of a student having at least one but not more than 2 siblings

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The probability of a student having at least one but not more than 2 siblings = P(1) + P(2)

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P(1) = number of students with 1 sibling / total number of students

From the given parameters, we have that:

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P(1) = \frac{18}{40}

Solving for P(2)

P(2) = number of students with 2 siblings / total number of students

From the given parameters, we have that:

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P(2) = \frac{10}{40}

P(1) + P(2) = \frac{18}{40} + \frac{10}{40}

Take LCM

P(1) + P(2) = \frac{18 + 10}{40}

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Divide numerator and denominator by 4

P(1) + P(2) = \frac{7}{10}

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Convert to percentage

P(1) + P(2) = 70\%

<em>Hence, the required probability is 70%</em>

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