Find the distance between points P(1, 2) and Q(4, 8) to the nearest tenth. i neeeeeed help

1 answer:

5 0

The answer is
the vector PQ has (4-1, 8-2) as components, so it is vecPQ= (3, 6)
using the definition
the distance PQ= sqrt ( 3²+6²)= sqrt(9+36)=sqrt(45)

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Answer:

The answer is below

Step-by-step explanation:

When computing quartile and decile, the data must be arranged in ascending order.

Given the data points:

13 13 13 20 26 26 29 31 34 34 35 35 36 37 38 41 41 41 42 43 46 47 48 49 53 55 56 62 67 82

The numbers are arranged in ascending order. The total number of terms is 30.

$First \ quartile(Q_1)=\frac{1}{4}(n+1)^{th}\ term\\\\where\ n=number\ of\ terms. Hence:\\\\Q_1=\frac{1}{4}(30+1)=7.75^{th}\ term=average\ of\ 7th\ and\ 8th\ term=\frac{29+31}{2} =30\\\\Q_1=30\\\\Third \ quartile(Q_3)=\frac{3}{4}(n+1)^{th}\ term\\\\Q_3=\frac{3}{4}(30+1)=23.25^{th}\ term=average\ of\ 23rd\ and\ 24th\ term=\frac{48+49}{2} =48.5\\\\Q_3=48.5$

$Second \ decile(D_2)=\frac{2}{10}(n+1)^{th}\ term\\\\where\ n=number\ of\ terms. Hence:\\\\D_2=\frac{2}{10}(30+1)=6.2^{th}\ term=average\ of\ 6th\ and\ 7th\ term=\frac{26+29}{2} =27.5\\\\D_2=27.5\\\\\\Eight \ decile(D_8)=\frac{8}{10}(n+1)^{th}\ term=\frac{8}{10}(30+1)=24.8^{th}\ term\\=average\ of\ 24th\ and\ 25th\ term=\frac{49+53}{2} =51\\\\D_8=51$